if 



m 

■■ 

m 



.. 






it. 



■ ' I 




Class 
Book 



M4. 



Copyright^ . 



t90 3<u 



COPYRIGHT DEPOSIT. 



1 



B 



i>#/- < 



**k* £*,":*££ .AW*-,vi I v. . I ■ 







■ 

1 1 
I ■ 

mm ISP 

mi 



ft 






v ;*> v 



■ 



m 



.v.. 



■ ■ 



TECHNICAL MECHANICS. 



EDWARD R. MAURER, 

Professor of Mechanics in the University of Wisconsin. 



FIRS T EDITION. 
FIRST THOUSAND. 



NEW YORK : 

JOHN WILEY & SONS. 

London: CHAPMAN & HALL, Limited. 

1903. 






%o" 



0^ 



T*Mg USKAKY OF 
CON a HESS, 

Two Cot»iu# Reoravee 

' U '1908 



DLASs^xXn. Mo. 



Copyright, 1903, 

BY 

EDWARD R. MAURER. 



ROBERT DRUMMOND, PRINTER, NEW YORK. 



PREFACE. 



This work is not a pure or an applied mechanics, but a theo- 
retical mechanics for students of engineering, and in accordance 
with the usage of some German writers, the title "Technical 
Mechanics" has been given to it. 

On the theoretical side, practically all the subjects treated 
have a direct bearing in engineering problems. Little atten- 
tion was paid to experimentation, principally because the 
author's own students come to him after having completed a 
laboratory course in elementary mechanics. In general, the 
theory in each chapter has been grouped together and separated 
from the applications. The principal equations and formulas 
are set in bold-faced type. 

On the applied side, no attempt was made to present fully 
any one subject, as the analysis of trusses, friction, balancing 
of rotating systems, etc., for the object in view was the illus- 
tration of the use of principles of mechanics and not treatments 
of roof-trusses, friction, balancing, etc. However, a sufficient 
number and the proper kind of examples have been included, 
it is believed, to give to the student a working knowledge of 
the subject. 

In Statics, especially, a distinctive feature is the nearly 
coextensive use of graphical and algebraic methods. This is 
due to the author's opinion that it is unwise to present graphical 
and analytical statics as separate courses, at least to beginners. 
The treatments of composition and equilibrium of forces are 
separate. This arrangement is largely due to the author's 
desire to develop the conditions of equilibrium for the different 



iv PREFACE. 

classes of force systems consecutively and to get all the prin- 
ciples of equilibrium together. The examples on the applica- 
tion of the principles of equilibrium are set off in a separate 
chapter, number VI. No attempt was made to arrange them 
in accordance with the orders of Chapters II and V. On the 
contrary, such an arrangement was avoided, and the entire aim 
was to emphasize the fact that there is a general method for 
solving the ordinary problems of Statics proper and that it con- 
sists principally in "applying" conditions of equilibrium. 

Kinematics is treated mainly as a preliminary to Kinetics, 
but harmonic motion is discussed more fully than usual in 
works of this kind. 

In Kinetics, D'Alembert's Principle is used considerably. In 
spite of the fact that it has been described as clumsy and old- 
fashioned, the author believes that the use of the principle in 
all but simple cases is decidedly helpful, and besides its use 
makes graphical methods possible. 

As to "mass and weight," what has been called the physicist's 
usage is followed, that is, mass means quantity of matter and 
weight the Earth's attraction. The author retains, however, 
the familiar equation m = W + g, and as holding not only in 
gravitation systems of units but in all systems in which "unit 
force gives unit mass unit acceleration." Only such systems 
are used herein, and then the familiar equation F=ma always 
holds. This of course requires recognition of special units of 
mass in gravitation systems. The author has succeeded best 
with his classes in this matter by doing more than defining these 
units — he has been using a name for one of them, the "English 
engineers' unit" (equal to 32. 2 ± pounds). So strongly does^ 
he feel that names are helpful in this instance that he has ven- 
tured to put them into print. The names herein used are "gee- 
pound" and "geekilogram," denoting 32 . 2 ± pounds and 9 . 81 ± 
kilograms respectively. They were adopted for their descrip- 
tiveness and as better than "matt" and "ert," the only other 
terms proposed in this connection so far as the author is aware. 

For the information of any instructor who may consider 
using this book as a text, the following modifications and abridg- 
ments are pointed out: Chapters III and IV may follow Chapter 



PREFACE. v 

VI, Chapter V may be taken simultaneously with Chapter II, 
and Chapters III, IV, and XV, and parts of all other chapters 
may be omitted without serious derangement of the course. 

In his work the author consulted especially the books of 
Profs. L. M. Hoskins and John Perry. He is especially indebted 
to Prof. C. H. Burnside of this University for valuable assistance 
on Statics. 

Madison, Wis., September, 1903. 



TABLE OF CONTENTS. 



INTRODUCTION. 

ART. 

Nature of the Subject i 

Division of the Subject 2, 3 

Two Methods of Analysis 4 

STATICS. 

CHAPTER I. FORCE. 

1. Preliminary. 

Force Denned 5 

Action and Reaction 6 

Force at a Distance and Force by Contact 7 

Distributed and Concentrated Forces 8, 9 

Graphical Representation of a Force 10 

Notation 11 

2. Measurement of Force; Mass and Weight. 

Mass 12 

Units of Force 13 

Measurement of Force 14 

Weight 15 

3. Force Systems. 

Definitions 16 

Classification of Force Systems 17 

CHAPTER II. EQUIVALENCE OF FORCE SYSTEMS. 

1. Preliminary. 

Definitions 18 

The Principle of Transmissibility 19 

Graphical Composition of Two Concurrent Forces 20 

Algebraic Composition of Two Concurrent Forces 21 

Graphical Composition of Three Concurrent Non-coplanar 

Forces 22 

Algebraic Composition of Three Concurrent Non-coplanar Forces 

Mutually at Right Angles 23 

vii 



vin TABLE OF CONTENTS. 

ART. 

Resolution of a Force into Two Concurrent Components. ....... 24 

Resolution of a Force into Three Non-coplanar Forces 25 

Moment of a Force with Respect to a Point . . 26 

" Varignon's Theorem " 27 

Moment of a Force with Respect to a Line 28 

Couples 29, 30 

Resolution of a Force into a Force and a Couple 31 

2. Collinear Forces. 

Composition 32 

3. Coplanar Concurrent Non-parallel Forces. 

Graphical Composition 33~35 

Algebraic Composition 36 

4. Coplanar Non-concurrent Parallel Forces. 

Graphical Composition 37-40 

The Principle of Moments 41 

Algebraic Composition 42, 43 

5. Coplanar Non-concurrent Non-parallel Forces. 

Graphical Composition 44, 45 

The Principle of Moments 46 

Algebraic Composition 47 

Reduction of a System to a Force and a Couple 48 

6. Non-coplanar Concurrent Forces. 

Graphical Composition 49 

Algebraic Composition 50 

7. Non-coplanar Parallel Forces. 

Graphical Composition 51, 52 

The Principle of Moments 53 

Algebraic Composition 54 

8. Non-coplanar Non-concurrent Non-parallel Forces. 

The Resultant 55 

Graphical Composition 56 

Principle of Moments 57 

Algebraic Composition 58 

9. Theory op Couples. 

Equivalent Couples 59 

Composition of Couples 60 

Resolution of a Couple 61 

CHAPTER III. CENTRE OF GRAVITY AND CENTROID. 

1. Centroid of Parallel Forces. 

Centroid Defined 62 

Determination of the Centroid 63 

2. Centre op Gravity of a Body. 

Definition and General Formulas 64 



TABLE OF CONTENTS. ix 

ART. 

Moment of a Weight with Respect to a Plane 65 

Centre of Gravity Determined by Integration 66 

Centre of Gravity Determined Experimentally 67 

3. Centroids of Solids, Surfaces, and Lines. 

Centroid Defined 68, 69 

Moment of a Volume, Area, or Length 70 

Plane of Zero Moment 71 

Centroids of Simple Solids and Surfaces 72-80 

Centroids of Solids and Surfaces Consisting of Simple Parts 81 

Centroids of Solids and Surfaces Considered as Parts of Other 

Solids or Surfaces 82 

Centroids Determined by Integration 83-85 

Theorems of Pappus and Guldinus 86 

CHAPTER IV. ATTRACTION AND STRESS. 

1. Gravitation. 

Law of Gravitation 87 

Gravitation Constant 88 

Density 89 

Attraction at a Point or "Strength of Field" 90 

Attractions in Some Simple Cases 9i~97 

2. Electric and Magnetic Attractions. 

Laws of Electrostatic and Magnetic Forces 98 

Strength of Field 99 

Analogy between Electrical or Magnetic and Gravitational 

Attractions. 100 

Strengths of Field Due to some Simple Distributions of Elec- 
tricity and Magnetism 101 

3. Stress. 

Stress Defined 102 

Units for Expressing Stress 103 

Classification of Stresses 104 

Description of a Simple Stress 105 

Intensity of Stress 106 

Graphical Representation of a Simple Stress 107 

Centre of Stress 108 

A Uniformly Varying Normal Stress 109-1 12 

CHAPTER V. GENERAL PRINCIPLES OF EQUILIBRIUM. 

1. Preliminary. 

Definitions 113 

General Condition of Equilibrium of a System of Forces Applied 

to a Rigid Body 114 



x TABLE OF CONTENTS. 

ART. 

Equilibrium of a System of Forces Applied to a Non-rigid Body. . 115 
2. The Conditions of Equilibrium for the Various Classes of Force 
Systems. 

Collinear Forces 116 

Coplanar Concurrent Non-parallel Forces ... 117, 118 

Coplanar Non-current Parallel Forces 119 

Coplanar Non-concurrent Non-parallel Forces 120, 121 

Non-coplanar Concurrent Forces 122 

Non-coplanar Non-concurrent Parallel Forces 123 

Non-coplanar Non-concurrent Non-parallel Forces 1 24 

Summary of Conditions of Equilibrium 125 

CHAPTER VI. APPLICATIONS OF THE PRINCIPLES OF 
EQUILIBRIUM. 

1. Preliminary. 

Nature of the Problems 1 26 

•General Method of Solution 127 

2. Flexible Cords. 

Definitions 128 

Tension in a Cord 129 

Position Assumed by a Cord Sustaining Loads 1 30, 131 

Position Assumed by a Heavy Flexible Cord Suspended from 

Two Points 132, 133 

3. Tackle. 

The Pulley 134, 135 

4. Smooth Supports. 

Definitions 1 36 

Reaction of a Smooth Surface 137 

Pin Joint or Hinge 138 

5. Three Typical Problems on Coplanar Non-concurrent Forces. 

Problem 1 139 

Problem II 140 

Problem III 141 

6. Jointed Frames. 

Definitions 142 

The Pin Pressures 143 

General Direction for Solving Examples 144 

7. Jointed Frames — Continued. 

Kind of Frames Considered 145 

"Force or Stress in a Member" 146 

Method for Determining the Force or Stress in a Member 147 

Graphical Method for "Analyzing" Trusses 148-151 

8. Rough Supports; Friction. 

Definitions 152-155 



TABLE OF CONTENTS. xi 

ART. 

Laws of Friction 156 

Determination of Coefficient of Friction 157 

Friction Circle 158 

Cone of Friction 159 

Belt Friction 160 

"Forces in Space" and Miscellaneous. 

Examples Involving Non-parallel Non-coplanar Forces 161 

Miscellaneous 162 



KINEMATICS. 

CHAPTER VII. RECTILINEAR MOTION OF A PARTICLE. 

ART. 

I. Velocity and Acceleration. 

Specification of Position 163 

Space-time Curve 1 64 

Displacement 165 

Kinds of Rectilinear Motion 166 

Velocity 167-169 

Velocity -time Curve .- 170 

Velocity Increment 171 

Kinds of Non-uniform Motion 172 

Acceleration 173-175 

"Acceleration-time" and other Curves 176 

II. Important Special Motions. 

Uniform Motion 177 

Uniformly Accelerated Motion 178 

Simple Harmonic Motion 179, 180 

Motion of the Piston of a Steam-engine 181 

CHAPTER VIII. CURVILINEAR MOTION. 

I. Velocity and Acceleration. 

Specification of Position 182 

Space-time Curve 183 

Displacement 184 

Kinds of Motion 185 

Velocity 186 

Speed-time Curve and Hodograph 187 

Acceleration 188 

II. Resolutions of Velocities and Accelerations. 

Components and Resultant of Velocities or Accelerations Defined 189 

Axial Components of a Velocity 190 

Tangential and Normal Components of Velocity 191 



=«" TABLE OF CONTENTS. 

ART. 

Axial Components of Acceleration 192 

Tangential and Normal Components of Acceleration 193 

III. Relativity of Motion. 

Meaning of Relative Path, Displacement, Velocity, and Acceleration.. . 194 

Definitions 195 

Relation between the Velocities and Accelerations of a Point 196 

Meaning of Composition of Motions 197 

IV. Composition of Simple Harmonic Motions. 

Mechanism for Compounding Simple Harmonic Motions 198 

Composition of Two Collinear S. H. M.'s of Equal Periods 199 

Resolution of a S. H. M. into Two Components 200 

Composition of Many Collinear S. H. M.'s of Equal Periods 201 

Composition of Two S. H. M. 's in Lines at Right Angles 202 

Resolution of a S. M. H. into Two Rectangular Components 203 

Composition of more than Two S. H. M.'s not Collinear 204 

CHAPTER IX. MOTION OF A RIGID BODY. 

I. Translation. 

Translation Defined 205 

Motions of all Points of a Body in Translation are alike 206 

Velocity and Acceleration of the Body 207 

II. Rotation. 

Rotation Denned 208 

Angular Displacement 209 

Angular Velocity 210-212 

Angular Acceleration 213-215 

Velocity and Acceleration of any Point of a Rotating Body 216 

III. Any Plane Motion. 

Plane Motion Defined 217 

Angular Displacement 218 

Angular Velocity and Angular Acceleration 219 

Velocity and Acceleration of any Point of the Body 220 

Plane Motion Regarded as a Combined Translation and Rotation 221 

Instantaneous Axis (of no Velocity) 222 

Instantaneous Rotation 223 



KINETICS. 

CHAPTER X. MOTION OF A PARTICLE (RESUMED) AND OF A 
SYSTEM OF PARTICLES. 

I. Mass and Mass-centre. 

Quantity of Matter 224 

Mass 225 

Practical Determination of Mass 226 



TABLE OF CONTENTS. xin 

ART. 

Moment of Mass 227 

Mass-centre Denned 228 

Relation between Mass-centre and Centre of Gravity 229 

II. Motion of a Particle. 

Laws of Motion 230 

Quantitative Expression of the Second Law of Motion 231 

Kinetic System of Units 232 

Relations between Force Units and between Mass Units 233 

Relation between Mass and Weight of a Body 234 

Acceleration of a Particle Acted upon by Several Forces 235 

Equations of Motion of a Particle 236 

III. Motion of a System of Particles. 

Definitions 237 

D'Alembert's Principle 23S 

Component of an Effective System Along any Line 239 

Motion of the Mass-centre of any System of Particles 24c 

Moment of the Effective System About any Axis 241 

"Angular Motion" of a System of Particles 242 

CHAPTER XI. TRANSLATION OF A RIGID BODY (RESUMED). 

I. General Principles. 

Equations of Motion 243 

Resultant of the Effective System 244 

II. Applications. 

General Method of Procedure 245 

Kinetic Reactions 246 

Vibrations 247-251 

Kinetic Friction 252 

CHAPTER XII. ROTATION (RESUMED). 

I. Second Moments of Mass (Moment of Inertia, etc.) 

Occurrence of Second Moments 253 

Moment of Inertia 254 

Radius of Gyration 255 

Relations between Moments of Inertia and between Radii of Gyration 

with Respect to Parallel Axes 256 

Composite Bodies 257 

Experimental Determination of Moment of Inertia 258 

Product of Inertia 259 

Principal Axes 260 

il. General Principles. 

The Effective Forces 261 

Moment of the Effective System 262 

Equations of Motion 263 

Resultant of the Effective System 264, 265 



xiv TABLE OF CONTENTS. 

ART. 

III. Applications. 

Determination of the Motion 266 

Motion of Pendulums 267 

Torsion Balance 268 

Conical Pendulum 269 

Weighted Conical Pendulum Governor 270 

Kinetic Reactions 271-274 

Balancing of Rotating Bodies 275-277 

Pivot and Journal Friction 278 

CHAPTER XIII. ANY PLANE MOTION OF A RIGID BODY 

(RESUMED). 

I. General Principles. 

The Effective Forces 279 

Moment of the Effective System 280 

Equations of Motion 281 

Resultant of the Effective System 282 

II. Applications. 

Determination of the Motion 283 

Kinetic Reactions 284 

Rolling Resistance 285 

CHAPTER XIV. WORK AND ENERGY. 

I. Work. 

Work Denned 286 

Expressions for Work Done by a Force 287 

Sign of Work 288 

Unit of Work 289 

Work Diagram 290 

Work Done by Gravity upon a Body in any Motion 291 

Work Done by Concurrent Forces and by their Resultant 292 

Work Done by a Pair of Equal, Opposite, and Collinear Forces 293 

Work Done by a Body Against a Force 294 

II. Energy. 

Energy Defined 295 

Kinetic Energy Defined 296 

Kinetic Energy of a Particle 297 

Kinetic Energy of any System of Particles 298 

Potential Energy Defined 299 

The Amount of Potential Energy 300 

Potential Energy of a System not Always a Definite Quantity 301-303 

Localization of Potential Energy 304 

Other Forms of Energy 305 

III. Principles oe Work and Energy. 

Principles of Work and Kinetic Energy 306 

Principle of Work and Energy for Conservative Systems 307 



TABLE OF CONTENTS. xv 

ART. 

Conservation of Energy 308 

Principle of Energy for Machines 309-311 

IV. Applications. 

Computation of Velocity and Distance 312 

Train Resistance 313 

Friction Brakes 314 

Efficiency of Tackle 315 

Efficiency of a Mine-hoist 316 

CHAPTER XV. IMPULSE AND MOMENTUM. 

I. Impulse. 

Impulse of a Force whose Direction is Constant 317 

Impulse of a Force whose Direction Varies 318 

Component of an Impulse 319 

Moment of the Impulse of a Force 320 

II. Momentum. 

Momentum of a Particle 321 

Components of a Momentum 322 

Moment of Momentum 323 

Momentum of a System of Particles 324 

Moment of the Momentum of a System of Particles 325 

Momentum of a Rigid Body in Special Cases 326 

III. Principles of Impulse and Momentum. 

Principles for a Particle 327 

Principles for a System of Particles . 328 

IV. Applications. 

Computation of Velocity and Time 329 

Pressures Due to Jets 330 

Sudden Impulses 331 

Force of a Blow and Recoil of a Gun 332 

Collision or Impact 333 - 335 

Ballistic Pendulum and Centre of Percussion 336 



APPENDIX A. 

VECTORS. 

Scalar and Vector Quantities A 1 

Vector Denned A 2 

Addition of Vectors A3 

Negative of a Vector A 4 

Subtraction of Vectors A5 



xvi TABLE OF CONTENTS. 

APPENDIX B. 

RATES. 

ART. 

Kinds of Variable Quantities B i 

Rate of a Uniform Scalar B 2 

Rate of a Non-uniform Scalar B 3 

Sign of a Rate. B 4 

Unit of Rates B 5 

Rate of a Uniform Vector B 6 

Rate of a Non-uniform Vector B 7 

Descriptive Terms B8 



APPENDIX C. 

DIMENSIONS OF UNITS. 

Magnitude of a Quantity Ci 

Fundamental and Derived Units C2 

Dimensions of Units C 3 

Application of the Theory of Dimensions C 4 



APPENDIX D. 

SECOND MOMENTS OF AREA. 

y 

I. Moment of Inertia. 

Moment of Inertia Denned . D 1 

Units of Moment of Inertia D 2 

Radius of Gyration D 3 

Relations between Moments of Inertia and between Radii of Gyration. D 4 

Composite Areas D 5 

II. Product of Inertia. 

Product of Inertia Denned D 6 

Units of Product of Inertia D 7 

Zero Products of Inertia D8 

Relations between Products of Inertia D 9 

Composite Areas D 10 

III. Relations between Moments of Inertia with Respect to Axes 

Inclined to Each Other. 

General Equations D n 

Geometrical Constructions D 12, 13 

Principal Axes and Moments of Inertia D 14 



TECHNICAL MECHANICS. 



INTRODUCTION. 

i. Nature of the Subject. — Mechanics, broadly defined, is 
the science which treats of motion. It is a natural science, and 
not a branch of mathematics as the student is apt to infer from 
his observation that mathematics is extensively used in the 
subject. Its foundations consist in a few grand generalizations 
from experience, such as Newton's Laws of Motion; the super- 
structure consists in the deduction of the logical consequences 
of those generalizations. 

2. Division of the Subject. — The following divisions may 

be made: 

. x , T . . ( Kinematics 

(i) Mechanics , ^ .^ 

(Dynamics | g^^ 

Kinematics treats of methods for the description of motion. 
Dynamics deals with the circumstances of motions, such as their 
causes, etc. Kinetics embraces that part of dynamics which 
deals with variable motion, and Statics that part dealing with 
uniform motion.* 

The above classification and order is usually followed in 
modern works on pure mechanics, and is no doubt the logical 

one. 

( Rigid Solids 

/ \ ^ i_ • r Non-rigid Solids 
(,) Mechamcs of \ Uqui * 

[_ Gases 

The general principles are the same for these latter branches, 

but there are special ones and special methods for each branch ; 



* In a treatment of mechanics on the plan outlined, it would develop 
that the principles relating to bodies at rest are included in Statics. 



2 INTRODUCTION. 

thus it is convenient to treat them separately. The present 
volume deals with principles common to all these branches, but 
relates especially to the first one. It is the practice in most 
American schools to present the mechanics of non-rigid solids, 
liquids, and gases in separate courses, the first two under the 
titles of Strength of Materials and Hydraulics respectively, and 
the last in connection with Thermodynamics. 

3. Technical Mechanics. — By this term is meant a presenta- 
tion of the general principles of mechanics with special refer- 
ence to their application in the fields of engineering. 

Although the order of treatment outlined in the first division 
of the subject (art. 2) is the logical one, another is followed 
herein. This is the historical order: Statics, Kinematics, and 
Kinetics. Statics as presented also occupies a more important 
part than in the outline above, and is not limited by the pre- 
ceding definition; it deals especially with principles and appli- 
cations relating to bodies at rest. 

Since the logical order is not followed, several principles 
appear in the early pages which are not fully explained or justi- 
fied until later. 

4. Two Methods of Analysis. — They are called the graphical 
and the algebraical, or analytical. 

In the first method, the quantities under consideration are 
represented by lines and the analysis is wholly by means of 
geometrical figures. In calculations by this method leading to 
numerical results the figures are accurately drawn to scale. 
In the second method the quantities under consideration are 
represented by symbols, and the analysis is by ordinary algebra. 
Calculations are carried out by arithmetic. 

The graphical method finds its best application in statics, 
though it is advantageous in the applications of kinematics to 
mechanisms. Statics is often treated entirely by one of the 
two methods, such a treatment being known as graphical or 
analytical statics, as the case may be; in this book both methods 
are employed. Some discussions and solutions are given by 
both methods by way of comparison, the others by that method 
which is the more suitable. 



STATICS. 

CHAPTER I. 
FORCE. 

§ I. Preliminary. 

5. Force Defined. — Bodies act upon each other in various 
ways, producing different kinds of results. Those actions 
which influence the motion of bodies are now to be considered. 

Definition. — An action of one body upon another which 
changes or which exerted alone would change the state of the 
motion of the body acted upon is called a force. 

We say that a force is applied to a body, a force is exerted 
upon a body, a force acts upon a body, etc. The last expression 
is faulty and misleading; for, since force is the name for a cer- 
tain kind of an action, the expression might be rendered thus: 
an action acts upon a body. Now it is not an action which acts, 
but some other body. The expression criticised is, however, 
a current one and is used in this book. 

Our first notions about force are founded on our experience 
with forces exerted by or upon ourselves. From this experi- 
ence we have learned that a force has magnitude, direction, and 
place of application. 

6. Action and Reaction. — When one body exerts a force 
upon another, the latter also exerts a force upon the former, 
and the two forces are equal in magnitude but opposite in 
direction. This fact is often referred to as the principle of 
" action and reaction,'" the phrase being an abbreviation of 
Newton's third law of motion. By "action" is meant one of 
the forces, and by "reaction" the other. 

7. Force at a Distance and Force by Contact. — This is a 
classification for convenience and is probably not based on fact. 

Gravitational, electrical, and magnetic forces have been 

3 



4 FORCE. [Chap. I. 

called actions at a distance, the force between the bodies con- 
cerned being exerted without their contact and, as was formerly 
supposed, without any material connection between them. It 
is now known that the force between two electrified bodies 
depends upon the medium in which they are placed, which is 
proof that the medium has to do with the exertion of the force 
by one body upon the other. Even gravitation, it is be- 
lieved, is not a true action at a distance, and "the earth and 
a stone do not strictly draw each other together, but are 
pushed together by something which extends from one to the 
other." But apparently these forces are actions at a distance, 
and they will be so called. 

Pressure of the atmosphere upon any object, the force of a 
hammer blow upon a nail, etc., are forces by contact. The 
place of application of a contact force is the surface of contact 
between the bodies concerned. 

8. Distributed and Concentrated Forces. — This is a classifi- 
cation for convenience and is not in accord with fact. 

A distributed force is one whose place of application is a surface 
or a solid. For example, the water pressure on a ship, the place 
of application being the wetted surface; the attraction of the 
earth upon a stone, the place of application being the solid 
denned by the stone; etc. All forces are really distributed 
forces. 

A concentrated force is one of finite magnitude whose place 
of application is a point. Such a force is imaginary since no 
actual (finite) force can be applied at a point, but there are 
actual forces whose place of application is very small and which 
may be regarded as concentrated forces. The conception is 
useful in developing principles relating to actual forces. Thus, 
as just stated, many actual forces are practically concentrated 
and may be treated as such, and a distributed force is regarded 
as consisting of a great number of concentrated forces. 

The line of action, or action line, of a concentrated force is a 
line parallel to its direction and containing its point of applica- 
tion. Note the distinctions between the terms action line, 
direction, and sense. To illustrate, imagine a pull exerted upon 
a sled by means of a single cord. The action line of the force is 




§L] PRELIMINARY. 5 

the line determined by the taut cord, its direction is upward 
and to the right 30 with the horizontal, and its sense is upward, 
and to the right, not downward and to the left. 

9. Specification of a Concentrated Force. — A concentrated 
force is completely specified by its magnitude, direction, and 
application point. It is explained later that the effect of 
such a force applied to a rigid body does not depend upon its 
application point, but only on its magnitude, action line, and 
sense, which are therefore called the essential characteristics of 
a force (as regards rigid bodies). 

10. Graphical Representation of a Force. — Since a force 
has magnitude and direction, it is a vector quantity;* the 
magnitude and direction of a force may therefore be represented 
by a vector — the length of the vector repre- 
senting to some scale the magnitude of the - 
force, and the direction of the vector giving 
the direction of the force. If the force be 

concentrated, the vector may also represent > 

the action line, but it is convenient to represent FlG - *• 

the action line separately. Thus suppose that the irregular 
outline in fig. 1 represents a body to which a force is applied 
at P horizontally to the right. The vector AB represents 
the magnitude and direction of the force, and a horizontal line, 
indefinite in length through P, its action line. 

If the vector were drawn through P it would serve also to 
locate the action line, but representation by two lines is con- 
venient especially when many forces applied to the same body 
are under consideration. The part of the entire figure in which 
the body is represented and the action lines are drawn is called 
the space diagram, and the part in which the vectors are drawn 
is called the vector diagram. 

11. Notation. — In the graphical analysis, each vector and 
the corresponding action line will be marked by the same letters, 
a capital at each end of the vector and the same small letters, 
one on each side, at the action line as in fig. 1. Reference to a 
force in the text will be by capital letters; thus "the force AB" 

* See Appendix A for brief discussion of vectors. 



6 FORCE. [Chap. I. 

means one whose magnitude and direction are represented by 
AB and whose action line is ab. 

In the algebraic analysis, a force will be denoted by a capital 
letter, which will also stand for the magnitude of the force. 

§11. Measurement of Force; Mass and Weight. 

12. Mass. — By mass of a body is meant the quantity of 
matter in it. There are several independent units of mass in 
use; only two will be described here. 

The ' ' British unit ' ' is the quantity of matter in a certain 
piece of platinum deposited in the Office of the Exchequer, 
London ; this unit is called a pound. 

The ' ' French unit ' ' is the quantity of matter in a certain 
piece of platinum deposited in the Palais des Archives, Paris; 
this unit is called a kilogram. 

Copies, more or less exact, of these standards and multiples 
and submultiples of them are in common use especially for the 
measurement of quantity of matter in trade. 

13. Units of Force. — A gravitation unit of force is a force 
equal to the Earth's attraction on a unit mass. 

Gravitation units are not constant with regard to place, for 
the Earth's attractions on equal masses at different places are, 
in general, not equal; in fact, the attractions are in the ratio of 
the values of g in the formula 

£ = 32.0894 (1 +0.0052375 sin 2 /)( i — 0.0000000957 e) 

computed for the two places, / denoting latitude and e eleva- 
tion above sea level in feet. The extreme variation is between 
the attractions at a high elevation on the equator and at the 
pole; this variation is but 0.6 per cent, while for points within 
the United States the maximum variation is about 0.3 per cent. 
Ordinarily no account need be taken of the differences in the 
gravitation units of force employed at different places, for errors 
introduced into engineering calculations by this variation are 
practically always negligible. 

It is customary to call any gravitation unit of force and the 



§ II.] MEASUREMENT OF FORCE. 7 

unit mass on which it is based by the same name.* Correspond- 
ing to the above-described units of mass we have then 

the unit of force called a pound, which is a force equal to 

the Earth's attraction on a pound mass, and 

the unit of force called a kilogram, which is a force equal 

to the Earth's attraction on a kilogram mass. 

An absolute unit of force is one whose value is independent 
of time or place. Several such units are described later. 

14. Measurement of Force. — The lever balance is primarily 
a force-measuring device, and it is the one commonly used for 
measuring forces. The force to be measured is applied at one 
end of a lever or system of levers, and the Earth's attraction on 
a body of known mass, m pounds say, at the other, the mass 
being such that the two forces balance. If the ratio between 
two such forces (determined once for all by the balance-maker) 
is n, then the magnitude of the force being measured is nm 
pounds. As the reader knows, the balance-maker provides that 
the numerical value, nm, may be read off directly. It should be 
noticed that a lever balance measures forces in terms of a gravi- 
tation unit. 

By means of lever balances, engineers measure forces applied 
to materials in testing their strength, and usually also, with a 
slight indirection, the "brake resistance" when testing engines 
and other motors. 

The spring balance is another force-measuring device. The 
force to be measured is applied to the spring so as to stretch it 
merely. The force corresponding to each amount of stretch, 
within the range of the spring, having been determined by the 
maker, the magnitude of the force being measured may be in- 
ferred from the stretch. As the reader well knows, the maker 
provides that the numerical value of the force may be read off 
directly. In theory at least, forces measured with a spring 

* Such usage is apt to confuse the beginner, and, when necessary for 
clearness, we shall add an explanatory word; thus, 'pound mass' or 
' pound force.' The word ' second' is used in a closely analogous way. 
It is the name for two different units, one of angle and the other of time, 
and the latter is denned with reference to the first. To be clear, we 
often say second of arc or second of time, as the case may be. 



8 FORCE. [Chap. I. 

balance are measured in terms of a gravitation unit for the place 
at which it was graduated. For, consider the theory of the 
graduation of a spring balance: a body of known mass, a pound 
say, is suspended from it, and the position of the pointer on the 
spring is scribed on the scale ; the position is marked one pound 
(force), for it corresponds to a stretch due to a pound force; 
then other bodies whose masses are multiples or submultiples 
of the unit mass are successively hung from the balance and the 
scale is marked correspondingly. 

For measuring some forces the spring balance is better 
adapted than the other. Engineers use it to measure the pres- 
sure of the steam in a running engine, the pull of a locomotive, 
etc. 

15. Weight. — By weight of a body will be meant the Earth's 
attraction upon it. Weight as here defined is a force and must 
be measured in force units. It is customary to express weights 
in gravitation units. 

As before stated, the weight of any body changes slightly 
with change of its locality. However, if it is expressed in a 
gravitation unit the numerical value of the weight remains the 
same, for the relative changes in the weight and the unit are 
equal. An analogy is the measurement of the length of an iron 
rod by a standard of the same material at two different times, 
the temperature having changed during the interval. The 
length of the rod has changed, but the numerical value as deter- 
mined by the iron standard remains constant, for the relative 
changes in the lengths of the rod and standard are the same. 
If the weight of a body is expressed in an absolute unit, the 
numerical value will change just as the weight changes if the 
body be transported. 

'A lever balance, then, will give the same numerical value for 
the weight of a body at all places, while a spring balance, if 
sufficiently sensitive, will show the true variation in its weight. 



§111.] FORCE SYSTEMS. 



§ III. Force Systems. 

i 6. Definitions. — Any number of forces collectively con- 
sidered is called a system of forces or a force system. 

The forces of a system are called coplanar when their action 
lines are in the same plane, and non-co planar when they are not 
in a plane. 

The forces of a system are called concurrent when their action 
lines intersect at a point, and non-concurrent when they do not 
so intersect. 

The forces of a system are called parallel when their action 
lines are parallel, and non-parallel when they are not parallel. 

Force systems may be described in accordance with the above 
definitions, thus: concurrent systems, non-coplanar parallel 
systems, etc., according as the forces of the system are con- 
current, non-coplanar and parallel, etc. 

A system of two forces which are equal, parallel, opposite in 
sense, and have different action lines is called a couple. 

17. Classification of Force Systems. — A classification may 
be made in various ways. In Chaps. ] I and V the treatment is 
based upon the following classification: 

, n .J Parallel 

Concurrent j Non-parallel 

Coplanar 1 (Parallel 

Non-concurrent i f arajjei 

( Non-parallel 

( Concurrent 
Non-coplanar-^ ^ T ( po r oiipi 

Non- concurrent \ f arauei 1 
( Non-parallel 



CHAPTER II. 
EQUIVALENCE OF FORCE SYSTEMS. 

§ . Preliminary. 

i 8. Definitions. — Equivalent force systems are such as may 
be substituted for each other without change of effect. 

The resultant of a force system is the simplest equivalent 
system. The resultant of a system acting upon a rigid body 
consists of either one or two forces, as will be shown later. It 
follows from the above definitions that two equivalent systems 
have the same resultant. The components of a force are any 
forces whose resultant is that force. 

Composition of a force system is the process of finding a 
simpler equivalent system. Finding the resultant of a system 
is the most important case of composition. Resolution of a 
force system is the process of finding a less simple equivalent 
system. Finding components of a force is the most important 
case of resolution. 

19. Principle of Transmissibility. — The effect of a force 
applied to a rigid body is the same for all application points in 
its action line. This follows from the equations of motion of a 
rigid body (art. 242) } for they are independent of the application 
points of the applied forces. 

This principle may be roughly verified by experiment with 
the apparatus represented in fig. 2, which consists of a body 
suspended from two spring balances. 
The springs are elongated on account of 
the weight of the body, and if a force, 
as F, be applied at A , the springs will 



nJ^f suffer additional elongations which in a 

Fig. 2. way are a measure of the effect of the 

applied force. If the application point of F be changed to B 



*!•] 



PRELIMINARY. 



1 1 



or C, the spring readings will not change, hence the effect of F 
has not changed. 

20. Graphical Composition of Two Concurrent Forces. — 

The Parallelogram Law. — If two forces acting upon a rigid 
body be represented by OA and OB, then their resultant is 
represented by the diagonal OC of the parallelogram OA BC 
(see fig. 3). 



;*c 




Fig. 3. 




The Triangle Law. — If two concurrent forces acting upon a 
rigid body be represented in magnitude and direction by AB 
and BC, their resultant is represented in magnitude and direc- 
tion by the side AC of the triangle ABC (see fig. 4). 

The second law is obviously a consequence of the first, and 
the parallelogram law may be verified experimentally as follows : 
Set a drawing board in a vertical position and fasten a spring 
balance and smoothly working pulleys to it somewhat as shown 
in fig. 5. Next fasten three cords to a small ring, tie their 
loose ends to the hook 
of the balance and to 
two bodies whose 
weights are known 
and lay two cords 
over the pulleys as 
shown. 

The strings now 
exert three forces on the ring (F lf F 2 , and F 3 ) equal to the 
weights (W t and W 2 ) of the suspended bodies and the reading 
(S) of the balance, respectively. Since F 3 "balances" F t and 
F 2 , the resultant of F 1 and F 2 must be equal and directly opposite 
to F 3 . We have only to ascertain now whether a construction 
according to the parallelogram law gives a resultant equal and 




Fig. 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



opposite to F 3 . So lay off on the board OA and OB equal (by 
some scale) to F t and F 2 , and complete the parallelogram 
OABC. Then OC should (according to the law) represent the 
resultant of F x and F 2 , i.e., it should represent a force equal and 
opposite to F 3 and it will be found that it does. 



EXAMPLES. 

i. The large square (fig. 6) represents a board 3X3 feet 
upon which five forces are applied as shown. Determine com- 
pletely the resultant of the 4- and 5-lb. forces. 

2. Determine completely the resultant of the 7- and 8-lb. 
forces. 

Q 



,K 4 lbs. 





^R 



(») 




7 lbs. 



Fig. 6. 

21. Algebraic Composition of Two Concurrent Forces. — 

Let P and Q (fig. 7) be two concurrent forces and a the angle 
between their action lines. Of course there are two angles 
here; the one taken is that between the parts of the lines 
on the sides of their intersection toward which the arrows 
point. 

The lines AB and BC represent the magnitude and direction 
of P and Q respectively; then AC represents the magnitude and 
direction of their resultant. The action line, parallel to AC> is 
marked R Since the angle CBD = a, 



and 



AC 2 = AB 2 +BC 2 + 2 AB BC cos a, 
tan CAD = BC sin a/(AB+BC cos a). 



§1.] 



PRELIMINARY. 



13 



If R denotes the resultant and 6 the angle between R and P, 
then, since AC represents the value of R and the angle CAD 
equals 0, 

R 2 = P 2 +Q 2 +2PQ cos a (1) 

and 

tan d = Q sin a/(P + Q cos a). ... (2) 

Special cases: If a = 90°, R=(P 2 + Q 2 )* and tan d=Q/P. 
Describe the resultant if a = o°; if a = i8o°. 




EXAMPLES. 

1. Solve ex. 1, art. 20, by the formulas of this article. 

2. Compute the resultant of the 6- and 7-lb. forces, fig. 6. 

3. Show how the magnitude of the resultant of two forces 
P and Q changes as a changes from o° to 180 . 

22. Graphical Composition of Three Concurrent Non- 
coplanar Forces. — Parallelo piped of Forces. — If three non-co- 
planar forces acting upon a rigid body 
be represented by OA, OB, and OC, their 
resultant is represented by the diagonal 
OD of the parallelopipedO,4£C-L>. (See 
fig. 8.) 

Proof : According to the parallelo- 
gram law OC represents the resultant 
of two of the forces, OD represents the 
resultant of the third force and OC and 
given forces. 

From an inspection of the figure, it is plain that the magni- 
tude and direction of the resultant of three non-coplanar con- 
current forces is given by their vector sum. 

23. Algebraic Composition of Three Concurrent Non-Coplanar 
Forces Whose Action Lines are Mutually at Right Angles. — 
Denote the three forces and their resultant by P, Q, S, and R, 
and the acute angles between P and R, Q and R, and S and 
R, by 6 lt 6 2 , and 6 3 , respectively. Supposing that P, Q, and 5 
are represented by OA, OB, and OC of fig. 8, it is plain that 

R 2 = P 2 + Q 2 +S 2 

and 

cos 6 1 = P/R, cos 6 2 = Q/R, cos d 3 = S/R. 



Fig. 8. 
hence of the three 



14 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



EXAMPLES. 

i. Suppose that a force of 9 lbs. acts at A (fig. 6), perpen- 
dicularly to the plane of the board and outward. Determine 
the resultant of the 9-, 4-, and 7 -lb. forces. 

2. Change the sense of the 9-lb. force and solve ex. 1. 

24. Resolution of a Force into Two Concurrent Components. 
— It may be performed by applying the triangle or parallelo- 
gram law inversely. Thus, let it be required to resolve the 
force AB (fig. 9), applied at P. Draw from A and B two lines 
which intersect in any point C\ then AC and CB represent 
the magnitudes and directions of components of AB. The 
action lines of the components must intersect on ab, and their 
application points must be rigidly connected with P. For, if 
the two forces AC and CB be compounded, their resultant will 
be found to be AB. 





a c 

( a ) (b) ' (a) (b) 

Fig. 9. Fig. 10. 

The problem just solved is indeterminate, for, C being any 
point, there are an infinite number of solutions. If conditions 
are imposed upon the components, the resolution is more or less 
definite; for example, if in the above it had been specified that 
the components should be horizontal and vertical, there would 
be but one answer. 

Rectangular Components, or Resolved Parts. — An important 
case of resolution is that in which the angle between the com- 
ponents is 90 . Each is called a rectangular component, or 
resolved part, of the force. They can always be readily com- 
puted. From fig. 10, it is plain that 
the rectangular com- "] { the cosine of the 

( the. macrnit.iidp. nf ) 



com- 
ponent or resolved 
part of a force along 
any line 



1 _ j the magnitude of 
I I that force 



acute angle be- 
tween the force 
and that line. 



If the rectangular components of a force F are parallel to 
coordinate axes, as x and y f they are called x- and ^-components 



§1.] 



PRELIMINARY. 



x 5 



of F respectively, and will be denoted by F x and F y . If the 
acute angles between the force and the x and y axes be denoted 
by a and 3 respectively, 



and 



F cos a 



F y = F cos p = F sin a. 



EXAMPLES. 

i. Resolve the 5-lb. force of fig. 6, page 12, into two com- 
ponents whose action lines are parallel to the 4- and 6-lb. 
forces respectively. 

2. Resolve the 5-lb. force of fig. 6 into two components 
whose action lines are parallel to the 8- and 6-lb. forces. 

3. Resolve the 8-lb. force of fig. 6 into two components one 
of which is horizontal and the other 6.5 lbs. in magnitude. 

4. Resolve the 6-lb. force of fig. 6 into two components 
whose lines of action are horizontal and vertical respectively. 

5. Resolve the 4- and 7 -lb. forces of fig. 6 into horizontal and 
vertical components. 

25. Resolution of a Force into Three Non-coplanar Forces. — It 
may be performed by applying the parallelopiped of forces 
inversely. Thus, let it be required to resolve the force repre- 
sented by OD (fig. 11). Construct a 
parallelopiped of which OD is a diag- 
onal. The three edges intersecting at 
O represent the components of OD; the 
application points must be rigidly con- 
nected to that of the given force. 

The problem just solved is indeter- 
minate, for any number of such paral- 
lelopipeds may be thus drawn. If 
conditions are imposed upon the com- 
ponents, the resolution is more or less 
definite. 

Rectangular Components. — An important case of resolution 
is that in which the three components are mutually at right 
angles. Each is a rectangular component, for the two com- 




Fig. II. 



16 EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

ponents of OD (fig. n), one of which is either OA, OB, or 0C> 
are at right* angles to each other, as OA and OA'. 

If the three components of a force F are parallel respectively 
to coordinate axes x, y, and z, they are called %-, y-, and 2-com- 
ponents of F, and will be denoted by F x , F y , and F z respec- 
tively. If the acute angles between the force F and the x, y, 
and z axes be denoted by a, /?, and y respectively, 

F x = Fcoso:, F 2/ =Fcos/?, F z = Fcos^. 

In some instances, it may be more convenient to determine these 
components as follows: First resolve the given force into two 
rectangular components one of which is parallel to one of the 
axes; the other will be parallel to the plane of the other two 
axes. Then resolve the second component into two forces 
which are parallel to these two axes. For example, if OD 
(fig. n) be resolved first along the x axis, the first resolution 
gives OA and OA' \ the resolution of OA' gives OB and OC. 
Also, 

OA = OD cos a, or F x = F cos a ; 

OB = OA' cos /?' = OD sin a cos /?', or F y = F sin a cos /?' ; 

OC = OA' cos j' = OD sin a cos f, or F Z =F sin « cos 7-'. 

26. Moment of a Force with Respect to a Point. — The 

moment of a force with respect to a point is the product of the 
magnitude of the force and the perpendicular distance between 
its action line and the point. The perpendicular distance is 
called the arm of the force with respect to that point, and the 
point is called an origin or a centre of moments. 

In the following, the moment of a force will usually be de- 
noted by M, and the moment of a force with respect to an origin 
by M . 

The moment of a force with respect to a point is a measure 
of its tendency to rotate the body upon which it acts about that 
point. For, if the body is fixed at that point but free to turn 
about it in a given plane, any force in that plane will cause it 
to rotate about the fixed point. The amount of this tendency 
is proportional to the magnitude of the force and to its arm with 



§ I.] PRELIMINARY. 17 

respect to that point, and hence to the moment of the force with 
respect to the point. 

The Unit Moment. — From the definition of moment of a 
force, it follows that the unit moment is the moment of a unit 
force whose arm is a unit length ; hence there are many units of 
moments. We have no short names for any of them, but they 
are called a foot-pound, inch-ton, etc., according as the units of 
length and force are the foot and pound, or inch and ton, etc. 

The Sign of a Moment. — Sign, plus or minus, is given the 
moment of force according as it produces or tends to produce 
counter-clockwise or clockwise rotation about the origin of 
moments. In the following, the rotation is supposed to be 
viewed from the reader's side of the printed page. 

27. "Varignon's Theorem." — The algebraic sum of the 
moments of two concurrent forces with respect to an origin in 
their plane equals the moment of their o 
resultant with respect to that origin. /I V# ^fc 

Proof: Suppose that the vectors marked I \ \/ 2r'/ ! 
P and Q and R (fig. 12) represent two /"TT/u^K. / ' 
concurrent forces and their resultant I / A&?' D /\ I 
respectively. Call the arms of the three v^rttitS/ / — / 
forces with respect to 0, p, q, and r and ^^ ^ ^ ^~ — V 
the angles between the action lines and a FlG - I2 - 

perpendicular to OA, a, /?, and 6 respectively. From the 
figure, it is plain that 

P cos (x+Q cos ft = R cos 6 ; 
therefore P OA cos a +0 OA cos ^ = R OA cos 6, 

or Pp + Qq.= Rr. q.e.d. 

Supply proof when is between P and R, or Q and R. 

According to the theorem, the moment of a force equals the 
sum of the moments of its % and y components ; often it is easier 
to compute this sum than the moment directly. When one 
component passes through the origin, the moment of the other 
equals that of the force. 



i8 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



EXAMPLE. 

Compute the moment of the 6-lb. force (fig. 6) about C 
directly and from its horizontal and vertical components. 

28. Moment of a Force with Respect to a Line. — If a foice 
be resolved into components parallel and perpendicular to a 
given line, the product of the magnitude of the perpendicular 
component and the distance from its action line to the given 
line is called the moment of the force with respect to the line. 
The line is called an axis of moments, and the distance referred 
to above is called the arm of the perpendicular component with 
respect to that axis. 




Fig. 13. 

The value of the moment of a force with respect to a line does 
not depend upon the point of the action line at which the force is 
resolved. For if resolved at different points, the perpendicular 
components will be equal and their arms also. See fig. ij, which 
represents a force acting in A'D resolved into components AB and 
AC at A and A'B' and A'C at A'; evidently AB and A f B f are 
equal and their arms, a and a' ' , also. 

The moment of a force with respect to a line is a measure of 
its tendency to rotate the body on which it acts about that line. 
For, if a body is free to turn about the line, Oz say, any force, 
as AD, acting upon it will cause it to rotate. Now the com- 
ponents of AD would produce the same effect upon the body as 
does AD, but a component parallel to the axis would produce 



§1.] PRELIMINARY. 19 

no rotation and therefore the rotation effects of AD and the 
perpendicular component, AB, are the same. But the ten- 
dency of AB to produce rotation is proportional to AB and its 
arm, hence to their product, or the moment of AD. 

The sign of the moment of a force with respect to a line is 
taken as positive or negative according as the force produces 
counter-clockwise or clockwise rotation about the axis of mo- 
ments. The sign will depend upon the side from which the 
rotation is viewed. If the axis is also a coordinate axis, the 
rotation is customarily viewed from its positive end. 

Proposition I. If a force be resolved into three rectangular 
components, its moment with respect to a line parallel to one 
of the components equals the sum of the moments of the other 
two with respect to that line. 

Proof: With respect to the z axis (fig. 13), for instance, the 
moment of F equals the moment of its component AB about the 
point P. The moment of AB about P equals the algebraic 
sum of the moments of its components, F x and F y , about P 
(Varignon's Theorem). But the moments of F x and F y about 
P are also their moments about the z axis, hence, etc. 

Proposition II. If a force be resolved into two components 
one of which intersects an axis of moments, the moment of the 
force with respect to the axis equals the moment of the other 
component. 

Proof when the components are rectangular: Suppose that 
AD (fig. 13) is the force, / the axis of moments, and AE the 
action line of the component which intersects /; then the other 
component is F x . According to Prop. I, the moment of the 
force AD, or F, about I equals the sum of the moments of F x 
and F y , F z being parallel to I. Since the moment of F y is zero, 
the moment of F equals the moment of F x , q.e.d. 

EXAMPLE. 

Compute the moments of each force of fig. 38 (c) with 
respect to the x, y, and z axes, the edges of the cube being 4 ft. 
long. 

29. Couples. — Definitions. — Two equal and opposite forces 
not collinear are called a couple. By arm of a couple is meant 



20 EQUIVALENCE OF FORCE SYSTEMS. [Chap. li- 

the distance between the action lines of its forces. The moment 
of a couple with respect to a point in its plane is the algebraic 
sum of the moments of its forces with respect to that point. 

By the definition, the sign of the moment of a couple must 
be the same as the sign of the algebraic sum of the moments of 
its forces. The sign of the sum is the same for all origins (see 
proposition below) and can be seen at a glance for an origin 
between the forces or on the action line of one of them. The 
sense of a couple refers to the sign of its moment. We speak of 
positive and negative sense or counter-clockwise 
and clockwise senses. By aspect of a couple is 
meant the aspect * of its plane. 
F Proposition. — The moment of a couple is the 
same for all origins, and it equals the product of 
the magnitude of one of the forces of the couple 
and the arm. Proof : The moments of the couple 
of fig. 14 with respect to 0', 0" , and 0" f are respectively, 

-F-0 F A+F-O r B = F.AB, 




F-0"A+F-0"B=F.AB, 



and F>0" , A-F-0'"B = F'AB. 

Since 0', 0", and 0'" represent all possible origins in the plane, 
the proposition is proved. 

30. Graphic Representation of a Couple. — The moment of a 
couple, its aspect, and its sense may be represented by a vector. 
The length of the vector is made equal, by some scale, to 
the moment of the couple, it is drawn normal to the plane of 
the couple, and its arrow is made to correspond with the sense 
of the couple. This correspondence depends upon some arbi- 
trary rule, such as the following: the arrow on the vector points 
in the direction from which the rotation of the couple appears 
counter-clockwise. The vector thus representing a couple will 
for brevity be called the vector of the couple. 

For example, the couple in the upper face of the parallelo- 

* Aspect of a plane refers not to its position, but to its direction; it 
is conveniently specified by the direction of a line normal to it. 



I-] 



PRELIMINARY. 



21 




Vector scale: 1 in. = 100 ft. lbs. 
Fig. 15. 



piped of fig. 1 5 is represented by the vector AB or CD. Each 

vector also represents any other 

couple whose moment, aspect, and 

sense are the same as the moment, 

aspect, and sense of the upper couple; 

the couple in the lower face is such 

a one. 

It is shown in art. 59 that couples 
whose moments, aspects, and senses 
are the same are equivalent. It is 
therefore consistent to represent by the same vector all couples 
whose moments, aspects, and senses are the same. Further, it 
follows from art. 242 that the effect of a couple applied to a 
rigid body depends only upon its moment, aspect, and sense, 
which are therefore the essential characteristics of a couple. 
Hence we may say that a vector completely represents a couple, 
although it does not give the forces or arm of the couple nor the 
position of its plane. 

31. Resolution of a Force into a Force and a Couple. — 
Proposition. — A force may be resolved into a force acting 
through any arbitrarily chosen point and a couple. 

Proof: Let F (fig. 16a) denote the force to be resolved, and 
P the chosen point, a distant from F. Imagine two opposite 
forces equal and parallel to F introduced at P (fig. 166). Ob- 
viously the three forces of this figure are equivalent to the given 
force, i.e., they are components of it. But the three forces may 
be grouped into a force at P and a couple. 




Observe that the component force has the same magnitude 
and direction as the given force, and that the moment of the 
component couple is the same as that of the given force about 
the chosen point. 



22 EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

Since couples whose moments, aspects, and senses are the 
same are equivalent, the force F (fig. 16a) is equivalent to F 
of Fig. 1 6c and the couple there represented, provided that its 
moment equals F-a. 

EXAMPLE. 

Resolve the 8-lb. force (fig. 6) into one acting at B and a 
couple; into one acting at C and a couple. 

§ II. Collinear Forces. 

32. Composition.* — Let F lf F 2 , F 3 , etc. (fig. 17), denote the 
^^^^ . forces to be compounded. From art. 21 

< < < C — ;> > > it follows that the resultant of those 
forces of the system having the same 
sense is equal to their sum, that is, 

the resultant of F lt F 3 , F 5 , etc., or R', =F 1 +F 3 +F 5 + . . . , 

and 

the resultant of F 2 , F 4 , F 6 , etc., or R" ,=F 2 +F A -\-F Q + .... 

Also, the resultant of the system, or R, equals the difference 
between R f and R". Hence if R be given sign, positive or nega- 
tive according as it acts right or left, 

R=R'-R" = (F 1+ F 3 +F 5 + . . .)-(F 2 +F 4 +F 6 + . . .) 
= F 1 -F 2 +F 3 -F 4 +... 
R = 2T. 

The action line of R is of course the same as that of the given 
forces. 

§ III. Coplanar Concurrent Non-parallel Forces. 

33. Graphical Composition. — Let AB, BC, CD, and DE 
(fig. 18) be the forces to be compounded. By the triangle 
law (art. 20), compound AB and BC and replace them by 

* In the following articles on composition it is assumed that the 
force systems are applied to rigid bodies. 



§ III.] COP LAN AR CONCURRENT NON-PARALLEL FORCES. 



23 



their resultant AC; compound AC and CD and replace them 
by their resultant AD; finally compound AD and DE and re- 
place them by their resultant AE. By successive replacements, 
the given system has been reduced to a single force, AE, which 
is therefore the resultant sought. Notice that the lines AC, AD, 




(a) (b) 

Fig. 18. 



ac and ad are not necessary in a solution; they are used here 
only for demonstration purposes. 

34. Force Polygon. — The figure formed by drawing in suc- 
cession lines representing the magnitude and direction of the 
forces of any system is called a force polygon for those forces, 
or for the system. In fig. 18, ABCDE is a force polygon for 
the given system. Several force polygons can be drawn for any 
system, one for each possible order of drawing the lines; thus, 
for the system compounded in the preceding article twenty-four 
polygons can be drawn, no two alike. 

Proposition. — The algebraic sum of the components of any 
system of coplanar forces along any line equals the component 
along that line of a force whose magnitude and direction are 
represented by the line drawn from the begin- 
ning to the end of the polygon for the system. 

Proof: Let ABCDE (fig. 19) be the polygon 
for the system. The components of the forces 
along the line A'C are, in magnitude and direc- 
tion, A'B', B'C, CD', D'E', and that of a force 
whose magnitude and direction are AE is repre- 
sented by A'E'. From the figure, it is plain 
that A'E' equals the algebraic sum of A'B' , B'C, CD', and 
D'E'. 




E' 

Fig. 



B' D' 
19. 



C 



24 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



35. Rule for Composition. — Draw a force polygon for the 
forces; then a line from the beginning to the end of it. That 
line represents the magnitude and direction of the resultant; 
its action line passes through the common point of the action 
lines of the given forces. 

Examining fig. 18, it will be seen that the vector sum of the 
given forces represents the magnitude and direction of the 
resultant. 

EXAMPLES. 

i. Determine the resultant of the forces represented in 
fig. 6. 

2. Solve the preceding example, taking the forces in a differ- 
ent order in the force polygon. 

36. Algebraic Composition. — Let fig. 20(a) represent the 
forces to be compounded and the body to which they are applied. 
Resolve each force at the origin into its x- and ^-components, 
and replace it by them; the resulting system is represented in 




Fig. 20. 

fig. 20(6). Next compound all the ^-components and replace 
them by their resultant, 1F X , and compound the ^-components 
and replace them by their resultant, 2F y \ the resulting system 
is represented in fig. 20(c). Now these three systems are equiva- 
lent and have, therefore, the same resultant. If R denotes the 
resultant and 6 its direction angle, from fig. 20(d) it is plain 
that 



i 



cos 6 = IF X /R and sin = iT^/R, 

and the action line of R contains the common point of those of 
the given forces. 



§IV.] COPLANAR NON-CONCURRENT PARALLEL FORCES. 



25 



Referring to the figure, or to prop., art. 34, it is plain that 
the resolved part of R along any line *F a f 9 . 

equals the algebraic sum of resolved parts 
of its components along the same line, that 
is, 

Rcos^ = 2'F x . 

EXAMPLES. 

i. Solve ex. 1 of the preceding article 
algebraically. 

2. Let F lt F 2 , etc., fig. 21, equal 8, 4, IG ' 2I * 

6, 12, 7, and 5 lbs. respectively, and compute their resultant. 




§ IV. COPLANAR NON-CONCURRENT PARALLEL FORCES. 

37. Graphical Composition. — Let AB, BC, CD, and DE 

(fig. 22) be the forces to be compounded. ABCDE is a force 




,"*7 



>?0 



(b) 

Fig. 22. 

polygon for the given forces, and O is any arbitrarily chosen 

point. 

First, resolve AB into AO and OB, 

BC ". BO " OC, 

CD " CO " OD, 
and DE " DO " OE. 

Next replace each given force by its components. By art. 24, 
the action lines of AO and OB may intersect anywhere on ab, 

11 11 a B Q a QQ « « c< a ^ 

•• « co «< OD " " " " cd, 

" " " " DO " OE " " " " <te. 



26 EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

It is therefore possible to choose the action lines of the com- 
ponents so that those of OB and BO, OC and CO, and OD and 
DO coincide. Having so taken them, it is plain that OB and 
BO, OC and CO, OD and DO, balance; therefore the system 
of components reduces to AO and OK. Finally, the resultant 
of AO and OE is AE (art. 20), which is also the resultant of 
the given system. 

38. Funicular Polygon, etc. — The point (fig. 22) is called 
the pole of the force polygon. Lines OA, OB, OC, etc., from 
the pole to the vertexes of the force polygon are called rays. 
Lines oa, ob, oc, etc., are called strings, which, considered collect- 
ively, are called a string, or funicular polygon. 

If the notation for graphical statics (art. 11) be used, the 
following rules for drawing the funicular polygon will be found 
helpful, but the beginner should not depend entirely on them. 

(a) The two strings intersecting on the action line of any 
force are parallel to the rays drawn to the ends of the vector 
corresponding to that force; or, the strings intersecting on mn 
are om and on. 

(b) A string joining points in the action lines of two forces is 
parallel to the ray drawn to the common point of the vectors 
corresponding to those forces; or, the string joining points on 
Im and mn is parallel to OM. 

39. Rule for Composition. — Draw a force and a funicular 
polygon for the forces ; then a line from the beginning to the end 
of the force polygon, and a parallel one through the intersection 
of the first and last strings of the string polygon. The first line 
represents the magnitude and direction of the resultant, and 
the second its action line. (The first and last strings are those 
corresponding to the rays drawn to the beginning and end of the 
force polygon.) 

It is plain from fig. 22 that the vector sum of the given 
forces represents the magnitude and direction of the resultant. 

EXAMPLES. 

Fig. 23 represents a board upon which several parallel forces 
are applied. In each example below determine completely the 
resultant. 



§IV.] COPLANAR NON-CONCURRENT PARALLEL FORCES. 27 

1. The magnitudes of F x , F 2 , etc., are 40, 10, 30, 20, 50, and 
15 lbs. respectively. Solve twice, drawing two funicular poly- 
gons, starting them at different points. 

2. The magnitudes of F l9 F 2 , etc., are 20, 10, 30, 30, 50, 
and o lbs. respectively. Solve twice, drawing two funicular 
polygons using different poles. 

40. The Resultant when the Force Polygon Closes. — In that 
case, the beginning and end of the force polygon (A and E t 



< 






8 ft.-— 


v 


' > 


" i " ,< y> 













Fl F 2 ^3 h F 6 F 6 

Fig. 23. 

fig. 22) coincide; hence the first and last strings are parallel. 
The forces corresponding to these strings, and to which the 
given system is equivalent, are equal and opposite (AO and OE) 
hence they constitute a couple and can not be compounded. 

If the strings ao and oe should happen to coincide, then the 
forces AO and OE would balance; and their resultant, and hence 
that of the given system, is zero. Therefore 

A funicular polygon for a system whose force polygon 
closes may be open or closed; if open, the resultant is a 
couple, and if closed, the resultant is zero. 

It is worth noting that in the second case, force and funicular 
polygons closed, each segment of the funicular polygon is the 
action line of two forces. 

EXAMPLES. 

1. The magnitudes of F lt F 2 , etc. (fig. 23), are 20, 55, o, 15, 
10, and 30 lbs. respectively. Determine their resultant. 

Solution: The force polygon is ABCDEF (fig. 24), and 
since it is closed, the resultant is a couple. If we regard A as 
the beginning of the force polygon, the system compounds into 
two forces AO and OF whose action lines are ao and of respec- 
tively. They constitute the resultant couple; its arm is the 
dotted line of the space diagram. The arm scales 2.58 ft. and 



28 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. it 



the forces 67.9 lbs., hence the moment of the couples is 175.2 
ft. -lbs. The sense is clockwise. 

2. Solve the preceding example, taking the forces in a 
different order in the force polygon. 

41. The Principle of Moments. — The algebraic sum of the 
moments of any number of coplanar parallel forces with respect 



b b 



h>- 



e e 



h- d 



/. - -- 

<k-± o 

^ Scale: 1 in. =4 ft. 



/ 



^' Scale: lin.=401bs. 

^^ (b) 

Fig. 24. 

to any origin in their plane equals the moment of their resultant 
with respect to the same origin. 

Proof: Let the system be that represented in fig. 22. Re- 
membering 

that AO and OB are concurrent components of AB, 
" BO " OC " " *' " BC, etc., 

and recalling also Varignon's theorem, we may write 

moment of AB = moment of AO + moment of OB; 
moment of BC = moment of BO+ moment of OC; 
moment of CD = moment of CO + moment of OD ; 
moment of DE = moment of DO + moment of OE. 

Therefore 

JXmoments of AB, BC, CD, and DE) 

= ^(moments of AO, OB, BO, OC, CO, OD, DO, and OE). 



§IV.] COP LAN AR NON-CONCURRENT PARALLEL FORCES. 29 

But the moments of OB and BO, of OC and CO, etc., are equal 

and opposite, hence 

^(moments of AB, BC, CD, and DE) 

= moment of AO + moment of OE. 
If the resultant is a single force, 

moment of .40 + moment of 0E = moment of AE\ 
if the resultant is a couple, 

moment of .4 + OE = moment of the couple (see art. 29). 

Therefore, in either case, the sum of the moments of AB, BC, 
CD, and DE = the moment of their resultant. Proof may 
readily be extended to a system of more than four forces. 

42. Algebraic Composition. — I. If the algebraic sum of the 
forces is not zero, the force polygon does not close (art. 34); 
hence the resultant is a force (art. 37). If F x , F 2 , etc., denote 
the forces and R their resultant, 

R = 2T, 

and the sense of R is given by the sign of IF. 

The action line may be determined by means of the prin- 
ciple of moments; thus, if IM denotes the algebraic sum of 
the moments of the forces with respect to any origin 0, and 
a the corresponding arm of their resultant R, 

21^ = Ra, or a = i , M /R. 

The resultant must act on that side of O which will make the 
sign of its moment the same as that of IM . 

II. If the algebraic sum of the forces is zero, the force poly- 
gon closes (art. 34); hence the resultant is a couple (art. 40). 
According to the principle of moments, the moment of this 
couple equals the algebraic sum of the moments of the given 
forces about any point. 

EXAMPLES. 

i. Solve ex. 1, art. 39, three times, using each time a new 
origin of moments. 

2. Solve ex. 2, art. 39, twice, using different origins of 
moments. 

3. Solve ex. 1, art. 40. 



3° 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



43. Two Unequal Parallel Forces. — This is a common case to 
which the following special methods may be applied. Let P 
and Q denote the forces, P the larger, and R their resultant. 





, <r-iv->* c ->\--/- - 

'(a) (b) 

Fig. 25. 
When P and Q are alike in sense (see fig. 25a), 

R-P+Q, 

the sense of R is the same as that of the given forces, 
x = Qc/R and y = Pc/R. 
When P and Q are unlike in sense (see fig. 256), 

r=p-q: 

the sense of R is the same as that of the larger force, 

x = Qc/R and y = Pc/R* 

The student should prove the expressions for x and y and show 
that R is correctly represented in the figure, i.e., that its action 
line is between the forces in (a) . but beyond the larger force 
in (6). 

The following relations are sometimes convenient: 

From either figure Px = Qy } therefore 

PAC = QBC t or 

AC/BC-Q/P; 

hence the action line of the resultant of two parallel forces 
divides any secant intersecting their action lines into two seg- 

* It is plain from these equations that the smaller R is (P and Q 
nearly equal) the greater x and y are. As P and Q approach equality, 
they become more nearly equivalent to a couple; also R approaches zero, 
and its arm with respect to any origin in the body to which P and Q 
are applied approaches <*> . Hence we arrive at the conception that a couple 
is equivalent to a force of zero magnitude with an infinite arm. 



§V.] CO PLANAR NON-CONCURRENT NON-PARALLEL FORCES. 3 1 

merits which are inversely proportional to the magnitudes of 
those forces. Again, 

P/BC = Q/AC = (P + Q)/(BC + AC) = (P-Q)/(BC-AC), 

or 

P/BC=Q/AC=R/AB; 

hence the forces P, Q, and R are proportional to the distances 
between the other two. 

EXAMPLES. 

i. Determine the resultant of F 1 and F 2 of ex. r, art. 39. 
2. Determine the resultant of F 2 and F 3 . 



§ V. Coplanar Non-concurrent Non-parallel Forces. 

44. Graphical Composition. — First method. This consists 
in compounding two of the forces, then their resultant and a 
third force, that resultant and a fourth, etc., until the simplest 
equivalent system has been found. Thus, let A B, BC, CD, 




Fig. 26. 

and DE (fig. 26) be the forces of such a system. According to 
art. 20, 

the resultant of AB and BC is AC, 

" AC " CD " A£>, 

and " " " AD " DE " ,4£. 

Therefore AE is the resultant sought. 

It may happen that some of the intersections in the space 
diagram which are necessary do not fall within convenient 
limits. This is apt to occur when the action lines of the given 
forces are nearly parallel. In such cases it is practically im- 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



possible to determine the action line of the resultant, and the 
following method should be employed. 

Second method. This is the same as that for the composi- 
tion of coplanar non-concurrent parallel forces explained in 




Fig. 27. 

arts. 37, 38, and 39, which the student should read in connection 
with fig. 27. 

EXAMPLES. 

1. Let F lt F 2 , etc., in fig. 28 equal 8, 4, 6, 7, 12, and 5 lbs. 
respectively, and determine their result- 
ant.* 

2. Solve ex. 1, drawing a second fu- 
nicular polygon in the same space diagram 
beginning it at some other point. 

5. Solve ex. 1, choosing a new pole 
but employing the same space diagram. 

45. The Resultant when the Force Poly- 
gon Closes. — Just as in the case of parallel 
forces, the resultant is a couple. In fact 
the explanation in art. 40 may be read in connection with fig. 27. 

46. The Principle of Moments. — The algebraic sum of the 
moments of any number of coplanar forces with respect to any 
origin equals the moment of their resultant with respect to the 
same origin. 

The proof given in art. 41, read in connection with fig. 27, 
applies to this proposition. 

47. Algebraic Composition. — I. // the algebraic sum of the 
x and y-components of the forces are not both zero, the force 

* Use scales not less than 1 in. =2 ft. and 1 in. =2 lbs. 




§V.] COP LAN AR NON-CONCURRENT NON-PARALLEL FORCES. 33 



polygon for the system does not close (art. 34); hence the re- 
sultant is a force (art. 44). Since 

R X = IF X and R y = IF y , 
R=(W x 2 + W y 2 )\ 

sin 6 = IFy/R and cos = iT^/R, 
being the direction angle of R measured from the x axis. 

The action line of the resultant may be found by the principle 
of moments. If IM denotes the algebraic sum of the moments 
of the forces with respect to any origin , and a the correspond- 
ing arm of R, 

IM = R-a, or a = 2'M /R. 
The resultant must act on that side of which will make the 
sign of its moment the same as that of IM . 

II. // the algebraic sums of the x and y-components equal 
zero, the force polygon for the system closes (art. 34); hence 
the resultant is a couple (art. 45). According to the principle 
of moments, the moment of the couple equals the algebraic sum 
of the moments of the given forces about any origin. 

For a system of couples, IF x and IF y equal zero; hence 
their resultant is a couple, and its moment equals the algebraic 
sum of the moments of the given forces about any point. 

EXAMPLES. 

1. Solve ex. 1, art. 44, algebraically. 

Solution: It will be convenient to tabulate the computation. 



F 


a 


a 


F x 


Fv 


F-a 


8 lbs. 

4 " 

6 " 

7 " 
12 •' 

5 '• 




45° 

63° 25' 
90 

36° 52' 
14 2' 


2.00 ft. 
1. 41 " 

1.79 » 
4.00 " 
3.20 " 
2.91 " 


+ 8.00 
+ 2.83 
-2.68 
0.00 
-9.60 
+ 4.85 




+ 2.83 

+ 5-3& 

— 7.00 

— 7.20 

— 1. 21 


— 16.00 

- 5- 6 4 
-10.74 
+ 28.00 
-38.40 
+ 14.55 








+3.40 


— 7.22 


-28.23 



In the first column the force magnitudes are recorded, in the 
second the acute angle between each force and the x axis (taken 
horizontal), and in the third the arms of the force with respect 
to the selected origin (centre of the board); then in the fourth, 
fifth, and sixth columns the computed values of the #-compo- 



34 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



nents, the ^-components, and 
sums of these are 3.40 lbs., ■ 



the 
-7.1J 



moments. The algebraic 
lbs., and —28.23 ft .-lbs. 



respectively 

(1) 



Hence 
R = 



V3 



2 .40 + 7 2 .22 



7.98 lbs.; 

(2) the sense of R is downward to the right, the angle with the 
x-axis being « 



sin* 



7.22/7.98 = 64° 47' 




(3) the action line of R is to the right of the centre of the board 
a distance 28.23/7.98 = 3.54 ft. 

2. Solve the preceding example, choosing 
a different origin of moments. 

3. Let F lf F 2 , etc. (fig. 28), be 20, 14.14, 
22.36, 15, 25, and o lbs. respectively. Com- 
pute their resultant. 

4. Let F lt F 2 , and F 3 (Fig. 29) be 10, 20, 
and 30 lbs. respectively. Determine the re- 
sultant, the square being 4X4 ft. 
of a System to a Force and a Couple. — 



Fig. 29. 

48. Reduction 

Proposition. — Any system of forces can be compounded into a 
force acting through any arbitrarily chosen point and a couple. 

Proof: * Each force of the given system may be replaced by 
a force acting through the chosen point and a couple (art. 31). 
Suppose such a replacement made for each force; the resulting 
system consists of a concurrent one and a system of couples. 
But the resultant of the concurrent forces is a single force acting 
through the chosen point (art. 33 or 36), and the resultant of the 
couples is a single couple (art. 47). 

Computation of the force and the couple. Let F (fig. 30) 
be one of the forces of the given system 
and O the chosen point. The components 
of F are F' (applied at 0) and the couple 
FF" . Observe now that the component 
of F r along any line is the same in magni- 
tude and sense as that of F along the same 
line and that the moment of the couple, FF", is the same as that 

* This proof is for a coplanar system, the chosen point being in the 
plane of the forces. Proof for the general case is given in art. 55. 




§ VI. ] NON-COPLANAR CONCURRENT FORCES. 35 

of F about 0. Hence the algebraic sum of the components of all 
the forces of the concurrent system along any line is the same 
as that of the components of the given forces along that line, and 
the sum of the moments of the couples equals the sum of the 
moments of the given forces about 0. 

If R denotes the resultant of the concurrent system, and C 
that of the couples, 

R =(ir x 2 +ir y 2 )\ 

sin d = iTy/R, cos = I¥ x /R 
d being the direction angle of the action line of R , and 

C = 2'Fa. 

§ VI. NON-COPLANAR CONCURRENT FORCES. 

49. Graphical Composition. — Imagine a force polygon 
ABC . . . N drawn for the forces to be compounded; it is not 
of course a plane one. According to the triangle of forces, the 
resultant of the first two forces, R', is represented in magnitude 
and direction by AC. Likewise the resultant of R' and the 
third forces, i.e., the resultant of the first three forces, is repre- 
sented in magnitude and direction by AD, etc. Finally, the 
resultant of all the given forces is represented by the line AN, 
joining the beginning and the end of the force polygon. 

Hence the magnitude and direction of the resultant is 
represented by the vector sum of the given forces, or, other- 
wise stated, by the line drawn from the beginning to the end 
of the force polygon for the system. 
The action line of the resultant, of course, passes through the 
common point of the action lines of the given forces. 

Since the force polygon is not plane, it is practically neces- 
sary to represent it by projections. The line representing the 
resultant may then be determined by its projections, in the 
projections of the polygon. This may be further explained by 
the solution of an 

EXAMPLE. 

Suppose three forces F lt F 2 , and F 3 to act at a point as 
shown in Fig. 31(a). The horizontal and vertical projections 



36 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



of are marked r and 0" respectively, those of the vector 
representing F x are marked F/ and F 2 " , etc.* 

Solution: At any point A we begin to construct the polygon 
for the forces— A 'B'C'U is its horizontal and A"B"C"D" is its 
vertical projection. Hence the line AD represents the magni- 




Fig. 31. 

tude and direction of the resultant; the action line passes 
through 0. 

If the horizontal and vertical projections of the vectors 
representing the given forces be regarded as force systems, then 
obviously A'B'C'D' and A"B"C"D" are the force polygons for 
those systems respectively. 

50. Algebraic Composition. — Let F[ t F", F'", etc., denote 
the forces to be compounded. At the common point of 
their lines of action, resolve each force into three rectangular 
(x, y, and z) components and replace the forces by them. Next, 
compound separately all the x, y, and ^-components and re- 
place them by their resultants, IF X , IF y , and IF Z . Now these 
three systems, the given one, the system of components, and 
the three forces, IF ' x , IF y , and IF Z) are equivalent; they have, 
therefore, the same resultant. If R denotes the resultant, and 



* All horizontal and vertical projections are marked by primes and 
double primes respectively. 



§VIL] 



NON-COPLANAR PARALLEL FORCES. 



37 



6 X , 6 2 , and d 3 the direction angles of its action line, from the third 
system it is plain that 

(i) R=(lF x 2 + l¥ y 2 + I¥ 2 2 )\ 

(2) cosfl^iTs/R, cos0 a = 2T,,/R, cos 6 3 = 2T./R, 

(3) the action line passes through the common point of those 

of the given forces. 

§ VII. NON-COPLANAR PARALLEL FORCES. 

51. Graphical Composition. — The vectors F x and F 2 (fig. 32) 
represent two parallel forces, F/ 
and F 2 their projections on the xz 
plane, and F" and F 2 " those on the 
yz plane. Let R represent the re- 
sultant of the two forces, and R' and 
R" the projections of the vector R 
upon the plane xz and yz respect- 
ively. 

If Fi and F 2 be regarded as 
forces, R' represents their resultant 
for 




FJF 2 =CB/CA -C'B'/C'A' -F{/Fj t 

i.e., R' divides the line A'B' into segments inversely propor- 
tional to i 7 / and F 2 \ and obviously R , =F 1 , +F 2 . Similarly, 
if i 7 /' and F" be regarded as forces, R" represents their re- 
sultant. To find the resultant, then, of two parallel forces, 

Project the vectors representing them upon two planes 
parallel to the forces, and find the resultants of these pro- 
jections regarded as forces. These resultants are projec- 
tions of the resultant sought. 
This method may obviously be extended to the composition of 
more than two forces. It will now be illustrated by the solu- 
tion of an 

EXAMPLE. 

Suppose parallel forces of 14, 12, 16, and 8 lbs. applied to a 
body at points whose x and y coordinates are respectively (2, 4). 



38 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chai\ IL 



(3> 5)> (4' 2 )> an( i (6, 3), all in feet, and suppose that the third 
force acts in the negative and the others in the positive direc- 
tion. 

Solution: The vectors representing the forces are projected 




Fig. 33. 



on the yz and zx planes; thus a f b' and a"b" (fig. 33) are the 
projections of the 14-lb. force, b'c' and b ,f c" those of the 12-lb, 
force, etc. Each projected system may be compounded by 
the method of art. 37, one force polygon, as ABCDE, sufficing 
for the two systems. The funicular polygon in the xz plane 
determines aV, and that in the yz plane determines a"e" . The 
action line of the resultant sought then passes through the point 
P in the xy plane. The magnitude and sense of the resultant 
are given by AE. 

52. The Resultant when the Force Polygon Closes is a 
couple. For, obviously, the resultant of all the forces of the 
system but one is equal and opposite to that one; hence that 
resultant and the last force constitute the resultant couple. 
The couple is indeterminate, for it depends on which force of 
the system is omitted; accordingly, the system may be reduced 



§VIL] 



NON-COPLANAR PARALLEL FORCES. 



39 



by this method to as many different couples as there are forces 
in the system, but they are equivalent to each other. 

Let the system consist of F lt F 2 , F 3 , etc., and let R denote 
the resultant of F 2 , F 3 , etc.; then the resultant couple consists 
of F t and R. Imagine the vectors representing the forces to be 
projected on two planes and, as above, denote the projections 
on one plane by F/, F 2 ', F 3 ', . . . and R', and those on the 
other by F/', F 2 ", F 3 " , . . . and R" '. Evidently, the resultants 
of the systems F/, F/, F 3 , etc., and F/', F 2 ", F 3 " , etc., are 
couples, and the resultant of the first is F/, R' , and that of 
the second is F/', R" . 

It might of course happen that F 1 and R coincide. In that 
case the resultant of the given system would be zero ; and since 
F/ and R f and F/' and R" would also coincide, the resultants 
of the systems F/, F 2 , F 3 , etc., and F/', F 2 ", F 3 " ', etc., 
would be zero, and the funicular polygons for those systems would 
close (art. 40). 

53. The Principle of Moments. — The algebraic sum of the 
moments of any number of parallel forces with respect to a line 
equals the moment of their resultant with respect to that line. 

Proof: Let F x and F 2 (fig. 34) be two forces of the system. 




Fig. 34. 



R f their resultant, and OP tlie axis of moments. Imagine a 
plane which contains the axis and is parallel to the forces; and 
another which also contains the axis and is perpendicular to 
the first plane. Suppose that A, B, and C are the poinds in 
which F x , F 2 , and R f intersect the second plane, and cali the 



40 EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

angle between the forces and that plane a. Then the moments 
of the three forces with respect to OP are respectively 

{F 1 saia)AA , t (F 2 sin a)BB', and {R f sin ajCC' . 

Evidently, the resolved part of R normal to the second plane is 
the resultant of the normal components of F x and F 2 . Then, 
according to the principle of moments for coplanar forces, 

{R f sin a)OC = (F 1 sin a)OA + (F 2 sin a)OB. 

This equation multiplied through by sin (POB) becomes 

{R' sin a)CC' = (F x sin a)AA' + (F 2 sin a)BB' . 

That is, the moment of R' equals the sum of the moments of 
F x and F 2 . 

If F t and F 2 have unlike senses, a slight change in the proof 
is necessary, which the student can make. The extension of 
the proof to more than two forces is quite evident. 

For the exceptional case in which the resultant is a couple, 
the proposition still holds if we define the moment of a couple 
with respect to a line to be the algebraic sum of the moments of 
its forces with respect to that line. 

54. Algebraic Composition. — I. If the algebraic sum of the 
forces is not zero, the force polygon for the system does not 
close; hence the resultant is a force (art. 51). If F lt F 2 , etc., 
denote the forces and R their resultant, 



R = IF 



the sense of R being given by the sign of IF. 

The action line may be determined by the principle of 
moments; thus, if 2M X and 2M y denote the algebraic sums of 
the moments of the forces with respect to two axes, x and y, 
which are perpendicular to the forces, and a and b the corre- 
sponding arms of the resultant, R, 

2M x =±Ra and IM y = ±Rb, 
whence 

a=±IM,/R, b=±2 , M„/R. 



NON-COPLANAR, NON-CONCURRENT, NON-PARALLEL FORCES. 4* 

The resultant must act on such sides of the x and y axis that 
the signs of its moments are the same as those of IM X and IM y 
respectively. 

II. If the algebraic sum of the forces is zero, the force poly- 
gon closes; hence the resultant is a couple (art. 52). As ex- 
plained in the article referred to, the resultant couple is not 
determinate. A resultant couple can be readily found by com- 
pounding by the method under I all the forces of the system but 
one. That resultant and the omitted force constitute a re- 
sultant couple. 

EXAMPLES. 

1. Compute the resultant of five forces of 15, 12, 20, 16, and 
21 lbs., the first three acting in the positive z direction and the 
last two in the negative, the coordinates of the points in which 
they pierce the xy plane being respectively (2, 3), (4, —2), 

( 2 , 4), (3, - 1 ), and (°;°)- 

2. Include a force of 10 lbs. acting in the negative z direction 

at a point whose coordinates are ( — 8, 10), and compound. 

§viii. non-coplanar, non-concurrent, non-parallel 

Forces. 

55. The Resultant. — Proposition I. A system of non- 
coplanar, non-concurrent, non-parallel forces may be com- 
pounded into two forces, the action line of one being in, and 
that of the other normal to, any arbitrarily selected plane. 

Proof: Let the action lines of the forces of the system be 
extended till they pierce the selected plane.* This will be 
referred to as the plane n. At each of these points resolve the 
corresponding force into two components, one in the plane and 
the other normal to it. The given system may be regarded as 
replaced by two, a coplanar system (the components in n), and 
a parallel system (the components normal to n). In general, 
each of these systems compounds into a single force, but either 

* A force whose action line is parallel to the plane should be replaced 
by its equivalent, a force in the plane and a couple whose forces are 
perpendicular to it. (Art. 31.) 



42 EQUIVALENCE OF FORCE SYSTEMS. . [Chap. II. 

or both of them may compound into a couple (arts. 45 and 51). 
But, as shown in art. 42, a couple may be regarded as "a zero 
force with an infinite arm," and with this understanding the 
proposition holds in all cases. 

Since the two forces do not usually intersect, it is not 
possible, in general, to compound them into one force. They 
may therefore be properly called a resultant of the system. 
For different planes n we arrive at different pairs of resultant 
forces, but since they are equivalent to the same system they 
are equivalent to each other. We will denote the resultant 
forces in and normal to the plane by R t and R n respectively. 

Proposition II. A system of non-coplanar, non-concurrent, 
non-parallel forces may be compounded into a force acting 
through any arbitrarily selected point and a couple. 

Proof: Each force of the given system may be replaced by 
a force acting through the chosen point and a couple (art. 31). 
Suppose such a replacement made for each force; the result- 
ing system consists of a concurrent system of forces and a 
system of couples. But the resultant of the concurrent forces 
is a single force acting through the chosen point (arts. 49 or 
50), and the resultant of the couples is a single couple (art. 
60). We will denote this force and couple by R and C 
respectively. 

In general, R and C may be compounded into two non- 
parallel unequal forces. For, C may be replaced by an equiva- 
lent couple one of whose forces intersects R (see art. 59), and 
that force and R may be compounded into a force; since this 
last force will not be in the same plane with the other force of 
the couple these two forces cannot be compounded, and they 
may properly be called a resultant of the system. 

If the plane of C happens to be parallel to R, C and R 
may be compounded into a single force. For, C may be re- 
placed by an equivalent couple whose plane coincides with 
R; and, the forces of that couple and R being coplanar, their 
resultant is a single force. That force is the resultant of the 
system. 

In the following three articles, methods are explained for 
determining the resultants above discussed. 



NON-COPLANAR y NON-CONCURRENT, NON-PARALLEL FORCES. 43 

56. Graphical Composition. — The graphical method can be 
employed to determine Ri and R n , the method of art. 44 for R t 
and that of art. 51 for R n . Some elementary principles of 
descriptive geometry are employed for determining the inter- 
sections of the lines of action of the forces of the given system 
with the plane tz. The method will be illustrated by means 
of an 



EXAMPLE. 



Let there be three forces in the system, the horizontal pro- 
jections of the vectors representing them being F/, F 2 ', and F 3 ' 
(fig. 35), and the vertical projections F/', F 2 ", and F 3 ". 




a/b/c 
//a/d a r — 7o>/ 

' d "----V 

Fig. 35. 

The plane it is taken to coincide with the xy plane, the 
forces piercing it in the points P lt P 2 , and P 3 respectively. The 
components of the given forces in the plane n are represented 
by the vectors F/', F 2 ", and F 3 ", while the components normal 
to that plane act through the points P t , P 2 , and P 3 , and their 
magnitudes are represented by the projections of F/, F/, and 
F 3 ' upon the z axis. 

The resultant of the coplanar system is represented in 



44 EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

magnitude and direction by A'D' , A'B'C'D' being a force poly- 
gon for the system, and its action line is marked R t , 

The resultant of the normal system is represented in magni- 
tude and direction by AD, ABCD being a force polygon for 
the system. The funicular polygon for the projection of the 
normal system on the xz plane is oa', ob f , oc' ', and od' , and R n ' 
is the projection of R n on that plane. The funicular polygon 
for the projection of the normal system on the yz plane is 
oa" , ob", oc", and od", and R n " is the projection of R n on 
that plane. Hence the action line of R n pierces the plane 
at P. 

57. Principle of Moments. — The moment of the resultant * 
of any system of forces about a line equals the sum of the 
moments of the forces. 

This follows from the proposition that the sums of the 
moments of the forces of equivalent systems about any line 
are the same, which we now prove. 

Let I denote the line or axis of moments and imagine any 
plane containing it as the plane 7T. Each 
of the systems may be compounded into a 
resultant consisting of two forces, one in 
and one normal to the plane tz; since the 
resultants are identical,! R t and R n may 
denote the forces of each resultant. 
z Let F (fig. 36) be one of the forces of 

either system. The x axis is taken along 
the line / and the y axis in the plane tz] then the xy and tz 



* Here "moment of the resultant" means the algebraic sum of the 
moments of the forces of the resultant. 

t Proof : Suppose that they are not identical, and call the forces of 
the resultant of one system Rt' and R n ', and those of the other Rt" and 
Rn". Being resultants of equivalent systems, they are equivalent to 
each other; hence if the forces of one resultant be reversed, the four 
forces would balance, i.e., their resultant would be zero. Now the re- 
sultant of Rt' and i?/' reversed, is in the plane x, so call it Rt'"; that 
of Rn and R n " reversed is normal to that plane, so call it R n '" . But 
the resultant of Rt'" and R n '" cannot be zero unless each is zero, and 
that is impossible unless Rt' and Rt" and R n ' and R n " are identical. 



V 

r- A 


Jf 


~ ¥ 


-'' V F / 

X 



NON-COPLANAR, NON-CONCURRENT, NON-PARALLEL FORCES. 45 

planes coincide. F is shown resolved at A into two compo- 
nents (F t and F n ) in and normal to the plane it. According 
to Prop. II, art. 28, 

the moment of F about /=the moment of F n about I; 
hence 

I(mom. F) = 2(mom. F n ). 

But R n is the resultant of all the F n 's, and, according to art. 53, 

mom. R n about / = I(mom. F B ) = i'(mom. F); 

hence the sum of the moments of all the forces of either system 
equals that of R n , that is, the sums of the moments of the 
forces of the equivalent systems are equal. q.e.d. 

■ 58. Algebraic Composition. — Determination of R t and R n . 
Fig. 37 represents one force, F, of a system and its components 




Fig. 37. 

F x , F y , and F Zi also its components Ft and F n , the plane n 
being taken coincident with that of the x and y axes. From the 
figure, it is plain that F n equals F z and that the x- and y-com- 
ponents of F t are the same in magnitude and direction as those 
of F. It follows that 



(1) 



R n = IF z and R,=(JF, +2T/)*; 



(2) the sense of R n is given by the sign of IF Z ; 

(3) the angles which R t makes with the x and y axes are 

cos" 1 I¥ x /R t and cos" 1 I¥ y /R t . 



4 6 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



The action lines of R n and R t are determined by the principle 
of moments, thus: 

the arm of R n with respect to any line in the plane tz equals 
(the moment sum of the given forces with respect to that 
lme)/R n ; 

the arm of R t with respect to any line normal to n equals 
(the moment sum of the given forces with respect to that 
Vne)/R t . 

Determination of R and C. Let jF lf F 2t F 3 , etc., denote the 
forces to be compounded and (fig. 38a) the point through 
which R is to pass. The components of F 1 are the force F x ' at 
and the couple (F lf F t "); the components of F 2 are the force 
F 2 ' at and the couple (F 2t F 2 ") ; etc. 

\v 



X 



\y 



A 



M 



lb. 



•X 



w -- 1 10 lb. ' 



/ z 



(a) 



(b) 

Fig. 38. 



(C) 



It is plain that the components of F t ', F 2r etc., along any 
line are the same in magnitude and direction as those of F Xi F 2 , 
etc., respectively; hence 

cos t = iTs/R, cos d 2 = 2F y /R, cos 6 3 = JF^/R, 

IF X , IF y , and IF Z denoting the algebraic sums of the %,- y-, 
and ^-components of the given forces, and d lt d 2 , and d 3 the direc- 
tion angles of R (art. 50). 

Imagine the couple C resolved into three components whose 
planes are respectively perpendicular to the x, y, and z axes 
(art. 61) and denote them by C x , C y , and C z . Now the given 
system and the system R, C xt C y , and C 2 (fig. 386), have the 



fix.] 



THEORY OF COUPLES. 



47 



same resultant, and hence their moment sums with respect to 
any axis are equal. But the moment sums of R, C x , C y , and C z , 
with respect to the x, y, and z axes are C x , C y , and C z . Hence, 
if IM X , I My, and IM Z denote the moment sums of the given 
system with respect to the coordinate axes, 

C X = 1M X , C y = ZM y , and C Z = IM Z . 

Also, by art. 60, C = {lW x +IW y +IW z f and 

cos <j> x = ^Mz/C, cos (f> 2 = i'My/C, cos </> 3 = IM Z /C, 

</> lt 2 , and (f> 3 denoting the direction angles of the vector of C. 

EXAMPLES. 

1. Compound the four forces of fig. 38 (c) into a force at 
and a couple, the edges of the cube being 4 ft. long. 

2. Compound the four forces into two whose action lines are 
in and normal to the xy plane. 

§ IX. Theory of Couples. 

59. Equivalent Couples. — Proposition. — Two couples whose 
moments, aspects, and senses are the same are equivalent; or, 
otherwise stated, two couples whose vectors are the same are 
equivalent. 

Proof: I. The planes of the couples coincide. Let (Pp)* 
and (Qq) (fig. 39) be the two couples, F t (identical with Qi). 
F 2 , F 3 . . . a system of which (Pp) is 
the resultant. Then the vector sum 
of F lt F 2 ,F 3 , . . . F n is zero, and their 
moment sum about any point equals 

pp. 

It will now be shown that the 
couple (Qq) is also the resultant of 
the system F lt F 2 , F 3 , . . . by show- 
ing that the system can be compounded into two forces which 
are identical with the forces of the couple. One of those two 
forces is F t , and the other one is the resultant of F lt F 3 , . . . 
which call R. Since the vector sum of all the forces of the sys- 

* By " couple (Pp) " is meant one whose forces are P and arm is p. 
For convenience of designation, the forces of the couple are sometimes 
marked P : and P 2 ; then P x equals P 2 . 




Fig. 39. 



4 8 



EQUIVALENCE OF FORCE SYSTEMS. 



[Chap. II. 



tern is zero, R is equal and opposite to F v i.e., R is the same in 
magnitude and direction as Q 2 . If a denotes the arm of R with 
respect to an origin on Q lt then 

i?a = moment sum of F 2 , F 3 , . . . (for that origin) = Pp = Qq. 

Since R equals Q, a = q, i.e., R and Q 1 coincide. 

Finally, since (Pp) and (Qq) are the resultants of the same 
system, they are equivalent. 

II. The planes of the two couples are parallel. Let {Pp) 
and (Qq) (fig. 40) be the two couples. According to case I, 

the couple (Qq) can be replaced by 
any other in its own plane pro- 
vided that its moment and sense 
are the same as those of (Qq). Let 
(Ss) be such a couple, its forces 
being parallel to those of (Pp) 
a and one in the y axis. Imagine a 
system of parallel forces F x (iden- 
tical with St), F 2 , F 3 , . . . the re- 
sultant of which is the couple (Pp). 
The vector sum of the system is 
zero, and its moment sums with 
respect to the x and z axes equal the moments of its resultant 
(Pp) with respect to the same axes, which obviously are zero 
and Pp respectively. 

It will now be shown that the couple (Ss), and therefore 
(Qq), is also the resultant of F u F 2 , F v ... by showing that the 
system can be compounded into two forces which are identical 
with the forces of (Ss). One of these two forces is F v and the 
other is the resultant of F 2 , F 3 , . . . which call R. Since the 
vector sum of all the forces of the system is zero, R is equal 
and opposite to F v i.e., it is the same in magnitude and direc- 
tion as S 2 . Let a and c denote the arms of R with respect to 
the x and z axes respectively ; then 




Ra = moment sum of F 2 , F 3 
and 



Rc= " 
Hence a = o 



■T 2> F 3> ' * 



about the x axis = o 

" z " =Pp = Ss- 



and c = s; i.e., R and So coincide. 



fix.] 



THEORY OF COUPLES. 



49 



Finally, since (Pp) and (Qq) are the resultants of the same 
system, they are equivalent. 

60. Composition of Couples. — Proposition I. The resultant 
of any number of couples is a couple. 

Proof: Let AA'B'B and AA"B"B (fig. 41) be the planes of 




two of the couples. Replace the couples by equivalent ones 
(Ff x ) and (Ff 2 ) and so that a force of one couple shall ' ' balance " 
a force of the other; these balancing forces must lie in AB. 
The four forces are equivalent to two, F in A'B' and F in A"B n ', 
which clearly constitute a couple, the resultant of the two given 
couples. Similarly, the resultant of this resultant couple and 
the third couple is a couple, etc. 

The student should supply a proof for the case in which the 
planes of the couples are parallel. 

Proposition II. The vector of the resultant of any number 
of couples is the sum of the vectors of those couples '. 

Proof: Consider first two couples, those compounded above. 
Let CM and CN (fig. 41) be their vectors, then CO is the sum 
of CM and CN. (In fig. 416), the planes of the given couples 
are represented by their traces CC and CC" with a plane per- 
pendicular to AB.) It must be shown that CO is the vector of 
the resultant couple (Ff). 

By construction, ~CM/Ff 1 = CN/Ff 2) and the angles CNO 
and C'CC" are equal; hence the triangles CON and CC'C" are 
similar, and 



CO/f = CM/f x = CN/f 2 , or CO/Ff=CM/Ff 1 =CN/Ff 2 ; 



5© EQUIVALENCE OF FORCE SYSTEMS. [Chap. II. 

that is, the length of the vector CO represents the moment of 
the resultant couple to the same scale according to which CM 
and CN represent the moments of the given couples. Since 
CM is perpendicular to CO ', CO is perpendicular to C'C" \ i.e., 
the vector CO is normal to the plane of the resultant couple. 
From an inspection of the figures, it is plain that the arrow on 
the vector OC and the sense of the resultant couple agree in 
accordance with the rule of vector representation of couples 
(see art. 30). The proof is easily extended to more than two 
couples. 

Special Cases. — (a) Three couples whose planes are mutually 
at right angles. Let the three planes be taken as coordinate 
planes and call the couples whose planes are perpendicular to 
the x, y, and z axis C x , C y , and C z respectively, C their re- 
sultant, and v X} v y , v z , and v their vectors. Then 

v = (v x 2 + v y 2 + v z 2 )^; hence 
C = (Cs 2 +C/+C 2 2 )*. 

Also, if <j> lt (j) 2 , and </> 3 denote the direction angles of v, 

cos<f> 1 = v x /v i cos 4> 2 = v y /v, cos (j> 3 = Vg/v; 
hence 

cos^^Cx/C, cos <f) 2 = C y /C i cos </> 3 = C 2 /C. 

(b) Couples whose aspects are the same. The resultant is 
a couple whose aspect is the same as that of the given couples 
and whose moment equals the algebraic sum of the moments of 
the couples, a result reached in art. 47 for coplanar couples. 

61. Resolution of a Couple. — It follows from the preceding 
article that a couple may be equivalent to two or more couples, 
which are therefore components of that couple. Also, to re- 
solve a couple we have only to resolve its vector, the component 
vectors being the vectors of the component couples. 

The resolution of a couple into three components whose 
planes are mutually at right angles is an important special 
case. Let C be the couple to be resolved and v its vector; and 
denote the direction angles of the vector by a, /?, and y, the 



§IX.] THEORY OF COUPLES. 5 1 

coordinate planes having been taken to coincide with the 
planes of the desired component couples. Let C xy C y , and C z 
denote the component couples which are perpendicular to the 
x, y, and z axes respectively, and v x , v y , and v z the correspond- 
ing vectors. Then 

^ x =^coso:, v y =vcos{i t v z = vcosy 1 
hence 

C x = Ccosa, C„ = Ccos£, C, = Ccosjr. 



EXAMPLES. 

i. Hold this book, opened 150 , before you and imagine two 
couples whose moments are 50 and 70 ft .-lbs. to act in the planes 
of the right- and left-hand covers respectively, their senses being 
clockwise as viewed by yourself. On the supposition that the 
book is a rigid body, determine the resultant of the two couples. 

2. Imagine two couples whose moments are 40 and 90 ft. -lbs. 
to act in the front and right-hand side of the parallelopiped of 
fig. 15, their senses being clockwise and counter-clockwise re- 
spectively as viewed by yourself. Determine the resultant of 
the four couples. 



CHAPTER III. 
CENTRE OF GRAVITY AND CENTROID. 

§ I. Centroid of Parallel Forces. 

62. Centroid Defined. — A system of parallel forces having 
fixed application points possesses an important property which 
is now to be investigated. 

First consider a system of two forces, P and Q, with appli- 
cation points at A and B respectively, fig. 25 (a) or (b). In 
aft. 43 it is shown that BC/AC = P/Q; hence C, determined 
by the equation, is independent of the angle between A B and 
the action lines of P and Q. Therefore, if the body upon which 
the forces act be turned in any manner, the direction of the 
forces remaining unchanged, their resultant will always pass 
through the same point of the body. The point C may hence 
be regarded as the application point of the resultant for all 
aspects of the body. 

Consider next a system of more than two forces. 
Let F' t F'\ F'", etc., denote the forces; 

R' the resultant of F' and F" , and C its application point; 
R" the resultant of R f and F" f , and C" its application 
point; etc. 
If the body be turned, R' always passes through C' t and R", 
which is also the resultant of F', F", and F"' , always passes 
through C" . The point C" may therefore be regarded as the 
application point of the resultant of F', F" , and F'" for all 
aspects of the body. Extending this reasoning to the resultant 
of the first four force z of the system, then to the resultant of 
the first five, etc., one arrives at the conclusion that 

the resultant of any system of parallel forces having defi- 

52 



§!•] 



CENTROID OF PARALLEL FORCES. 



53 



nite application points always passes through the same 
point of the body, or its extension, irrespective of its aspect. 
This point is called the centroid of the system of forces. 

63. Determination of the Centroid. — Call the forces F', F" y - 
F'" , etc., those of one sense being given the same sign and those 
of the opposite sense the opposite sign. Let (V, y',z'), (x", y" , 
z"), etc., denote the coordinates of their respective application 
points with respect to a set of rectangular axes which is fixed 
with reference to the body, and let ~x, y, z denote the coordinates 
of the centroid. Now imagine the body upon which the forces 
act to be turned so that one of the axes, say that of x, becomes 

y 




parallel to the forces (see fig. 42). Then according to the 
principle of moments (art. 53), i^ being equal to IF, 



and 



Ry = F'y' + F"y" + F" , y ,n + . 



Now imagine the body turned until one of the other axes, that 
of y say, becomes parallel to the forces. The moment equation, 
with the z as moment axis, is 

Rx = F'x'+F"x"+F'"x'"+ .... 



From these three equations the desired formulas for the coor- 
dinates of the centroid follow; thus, 

x = IFx/R y y=2Ty/R, z = i'Fz/R. 



54 CENTRE OF GRAVITY AND CENTROID. [Chap. III. 

In these formulas signs must be given the forces as explained 
above, and to the coordinates of their application points as 
customarily. That this is necessary will be seen from an inspec- 
tion of the moment equations above and the figure. 

// the application points of the forces be coplanar, two of the 
formulas will suffice, provided that two of the coordinate axes 
be taken in the plane of the application points. The graphic 
method is easily applied in this case as follows : Imagine the 
body to be turned so that the action lines of the forces fall into 
the plane of the application points, their directions remaining 
unchanged. The force system is then coplanar, and the action 
line of its resultant may be readily determined graphically 
(art. 39). Now imagine the body to be turned about an axis 
which is perpendicular to the plane of the application points 
through any angle other than 180 or 360 degrees. The forces 
are still coplanar, and the action line of their resultant may be 
determined as before. The intersection of the action lines of 
the two resultants is the centroid of the system. 

EXAMPLES. 

i. Forces of 10, 20, 15, and 5 lbs. have the same directions; 
the application points are coplanar and their coordinates in feet 
are respectively, (5, 3), (2, 4), (1, 5), and (5, 6). Determine the 
centroid of the forces. Ans. x==2.6it. 

2. With the forces of the preceding example, include two of 
20 and 30 lbs. whose directions are the same as that of the four 
and whose application points are respectively at (5, 4) and 
( — 6, 3). Determine the centroid of the six forces. 

Ans. x = o.$ ft. 

3. Reverse the sense of the two additional forces, and deter- 
mine the centroid of the six. 

§ II. Centre of Gravity of a Body. 

64. Definition and General Formulas. — The weights of the 
particles of a body constitute a system of practically parallel 
forces having fixed application points. Therefore, these forces 
have a centroid, that is, the resultant always passes through a 
certain point fixed with reference to the body no matter how it 



§11.] CENTRE OF GRAVITY OF A BODY. 55 

is turned. It is assumed, of course, that in turning the body, 
its form and size are not changed. 

Definition. — The centre of gravity of a body is the centroid 
of the weights of all its particles. 

Centres of gravity, in the following, will be usually specified 
by means of rectangular coordinates and which will then always 
be denoted by x, y, and ~z. General values for these may be 
deduced as follows: Let (x', /, z'), (x",y", z"), etc., denote the 
coordinates of the particles* of a body; w f , w" , etc., their 
weights, and W the weight of the body. Then, from the 
formulas for centroid of a system of parallel forces, 

x = Iwx/W, y = 2wy/W, z = 2wz/W. 

65. Moment of a Weight with Respect to a Plane. — Defini- 
tion. — The moment of the weight of a body with respect to a 
plane is the product of the weight and the ordinate of the centre 
of gravity of the body from that plane. Ordinates on opposite 
sides of the plane are given opposite signs. A moment, there- 
fore, has the same sign as the corresponding ordinate. 

From the definition it follows that if the moment of the 
weight of the body is zero with respect to a plane, its centre of 
gravity is in that plane. 

Proposition. — The moment of the weight of a body with 
respect to a plane equals the algebraic sum of the moments of 
the weights of its parts with respect to the same plane. 

Proof: Let W lt W 2 , etc., be the weights of the parts and x/. 
x t " ', etc., be the ^-coordinates of particles of the first part, and 
w/, w x " , etc., their weights; x 2 , x 2 " , etc., be coordinates of 
particles of the second part, and w 2 , w 2 ", etc., their weights, etc 
Then, according to the formulas above, the ^-coordinate of the 
centre of gravity is 

__ (w 1 'x 1 ' + w 1 "x 1 " + . . .) + (w 2 'x 2 '+w 2 "x 2 " + . . ,) + etc. 
X ~ (w 1 '+w 1 " + . . ,)-fc («','+«/," + . . .)+etc 

* By particle is meant a body whose dimensions are vanishingly small 
or negligible in comparison with other distances involved, which, in this 
case, are for each particle the distances between it and the coordinate 
planes. 



5& 



CENTRE OF GRAVITY AND CENTROID. [Chap. III. 



Now the first parenthesis of the numerator equals W t x ly the 
second parenthesis equals W 2 ~x 2 , etc., and the denominator 
equals the weight of the entire body. The equation may there- 
fore be written 

Wx=W 1 x 1 +W 2 x 2 + . . . 

Since the plane from which the x's are measured may be taken 
at pleasure, the proposition is proved. 



i 
wire. 



EXAMPLES. 

Suppose that the heavy lines of fig. 43 represent a bent 
Determine the coordinates of its centroid. 

Ans. £ = 4. 13 in. 

Br 





Fig. 43- 



Fig. 44. 



2. A piece of tin consists of three parts, square, semicircular, 
and triangular, as shown in fig. 44. Determine the centre of 
gravity of the piece. (The distances of the centres of gravity 
of the semicircular and triangular parts from the base are 4/3^ 
times the radius and J times the altitude respectively.) 

3. Imagine the triangular and circular parts described in the 
preceding example bent forward on the lines BC and AC re 
spectively until their planes are perpendicular to that of the 
square. Determine the centie of gravity. 

4. Suppose that the triangular part of the preceding example 
is bent backward instead of forward. Write the expressions 
for the coordinates of the centre of gravity. 

5. The weights of four bodies are W iy W 2> W 3 . and W 4 , and 
the distance of the centre of gravity of the first from the plane 
through the centres of gravity of the other three is h How far 
from that plane is the centre of gravity of the four bodies? 

6. The weights of three bodies are 12, 18, and 40 lbs.; the 



§11.] CENTRE OF GRAVITY OF A BODY. 57 

distances between the centres of gravity of the first and second, 
second and third, third and first, are 10, 14, and 16 inches re- 
spectively. How far from the line joining the centres of gravity 
of the first and second is the centre of gravity of the three ? 

Arts. 7.91 in. 
7. Fig. 45 represents a cylindrical body having a cylindrical 
hole in the top; a part of the body is cast iron y.^ 5 .„ ..^ 
and the remainder (conical) is lead. Determine 
the centroid. (Cast iron and lead weigh 450 
and 711 lbs. per cu. ft. respectively. See art. 80.) 



66. Centre of Gravity Determined by Integra- 



tion. — Imagine a body divided into an infi- ! 



1 Void 1: ;■•'-.'-'• 

It J *— - J : ■ 

'Cast.'lto'h'.- 







nitely large number of parts, i.e., elements. 
Let dW denote the weight of any element and 
x, y, z, the coordinates of its centroid. Accord- 
ing to the principle of moments, 

Wx=/dWx, Wf = JdWy i Wz = fdWz. (1) 

If the body is homogeneous, let w denote its FlG - 45- 
specific weight * and V its volume ; then W = w V and dW =wdV. 
Equations (1) reduce to 

Vx = fdVx } Vy = ydVy, Vz = f&Vz. . . (2) 

These formulas may be employed for determining the centre of 
gravity of a body which cannot be divided into finite parts whose 
weights and centres of gravity are known, provided that its 
form and specific weight, if the body is not homogeneous, are 
such that the integrations can be performed. 

EXAMPLES. 

i. Determine the centre of gravity of an octant of a sphere 
whose specific weight, varying from point to point, is directly 
proportioned to distance from the centre. 

Solution: Let the plane faces of the octant be taken as 
coordinate planes as shown in fig. 46 ; denote the radius of the 

* By specific weight is meant weight per unit volume. 



58 



CENTRE OF GRAVITY AND CENTROID. [Chap. III. 



sphere by r and the specific weight at any point P by w. If k 
is a proper constant and x, y, and z denote the coordinates of P, 

w = k(x 2 +y 2 +z 2 )? and dW = k(x 2 +y 2 + z 2 )i dxdydz. 
Then from (i), 

Wx=kl J J (x 2 +y 2 + z 2 )? dxdydz- x; 

hence ~%=%r. Evidently, %=y=z. 




2. Determine the centre of gravity of an octant of a homo- 
geneous sphere. 

Solutions: Equation (2) maybe employed. First, selecting 
a cubical element as in ex. 1, dV = dx dy dz, and eq. (2) becomes 

pr pirt-xrf p(r2- x *—yrf 

Vx = / / / dx dy dz • x ; 

hence x = fr. Evidently, :v=37=2. 

Second, selecting the parallelopiped AB as elementary 
volume, dV = dx dy(AB). If # and y denote coordinates of B, 

AB = (r 2 -x 2 -y 2 )i, 
and eq. (2) becomes 

pr ptri-xrf 

I I {r 2 —x 2 — y 2 )^dxdy'X\ 



Vx = 



hence 



x=%r. 



§11.1 CENTRE OF GRAVITY OF A BODY. 59 

Third, selecting the volume between two planes parallel to 
the yz plane, dx apart as elementary volume, dV = nDC dx/4. 
If x denotes the coordinate of C, DC =r 2 —x 2 , and eq. (2) be- 
comes 

Vx=- f r (r 2 -x 2 )dx-x; 
hence x=%r. 

67. Centre of Gravity Determined Experimentally. — Some 
bodies are so irregular in shape that their centres of gravity can- 
not be found by the methods explained above. In such cases, 
experimental methods may be resorted to. 

The Method of Suspension. — The body whose centre of 
gravity is to be determined is suspended from one point of it and 
the direction of the suspending cord is then marked in some way 
on the body. The operation is repeated using another point of 
suspension. Since the centre of gravity is in each of the lines 
so fixed in the body, it is at their intersection. 

The Method of Balancing. — The body whose centre of gravity 
is to be determined is balanced upon a straight-edge and the 
position of the vertical plane containing the edge marked upon 
the body ; then the operation is repeated for two more balancing 
positions of the body. Since the centre of gravity is in each 
plane so fixed, it is in the common point of the three. 

This method is easily applied to determine the centre of 
gravity of a thin plate. The plate is balanced in two positions, 
same face down both times, and the lines of contact of the 
straight-edge and plate are marked upon the latter. The centre 
of gravity is midway between the intersection of those lines and 
a point directly opposite on the other face of the plate. 

EXERCISE. 

Cut from a sheet of stiff paper the ' ' angle section ' ' described 
in ex. 4, art. 81, and determine how far the centre of gravity of 
the paper is from the edges AC and CB. 



6o CENTRE OF GRAVITY AND CENTROJD. [Chap. III. 



§ III. Centroids of Solids, Surfaces, and Lines. 

68. Centroid Defined. — The term centre of gravity, applied 
to solids,* surfaces, or lines, is inappropriate, for they have no 
weight. However, the term is much used in that connection. 
Centroid, a more suitable term, is coming into use. Instead of 
the terms centroid or centre of gravity of a solid surface or line, 
centroid or centre of gravity of a volume area or length, re- 
spectively, are often employed. 

Definitions. — The centroid of a solid is that point of it which 
coincides with the centre of gravity of a homogeneous body which 
is bounded by the surface of the solid. The centroid of a surface 
is the limiting position of the centre of gravity of a homogeneous 
thin plate one of whose faces coincides with the surface as its 
thickness approaches zero. The centroid of a line is the limiting 
position of the centre of gravity of a homogeneous slender rod 
whose axis coincides with the line as its sectional area approaches 
zero. 

69. Centroid as Mean Point. — Proposition. — The ordinate 
of the centroid of any solid, surface, or line with reference to any 
plane equals the mean of the ordinates of all the equal elements 
of the solid, surface, or line. It must be understood that oppo- 
site signs are given to ordinates on opposite sides of the plane. 

Proof for lines: Let x t1 x 2 , x 3 , etc., denote the ordinates to 
the different elementary lengths ds of the line, and n the num- 
ber of elements (infinite). Then 

the mean ordinate = — , 

n 

, 1 (* i+*2+y 8 + ■••)<& f xds 
but this equals — 4 = — 1 — » 

I denoting the length of the curve. In art. 83 it is shown that the 
last expression is that for the ordinate of the centroid, hence the 
proposition is proved. 

Proof for solids and surfaces is similar to that just given. 



* Geometrical and not physical solid is meant. 



§111.] CENTROIDS OF SOLIDS, SURFACES, AND LINES. 61 

70. Moment of a Volume, Area, or Length. — Definitions. — 
The moment of the volume of a solid (area of a surface or length 
of line) with respect to a plane is the product of the volume 
(area or length) and the ordinate of the centroid of the solid 
(surface or line) from that plane. Ordinates on opposite sides 
of the plane are given opposite signs; hence a moment has the 
same sign as the corresponding ordinate. 

Proposition. — The moment of the volume of any number of 
solids with respect to a plane equals the algebraic sum of the 
moments of the volumes of those solids with respect to the same 
plane. Similar propositions hold for surfaces and lines. 

Proof for volumes: Let V lt V 2 , etc., denote the volume of the 
solids, u t , u 2 , etc., the ordinates of their centroids, and u that of 
the centroid of the collection of solids with respect to any plane. 
Now the ordinates of the centres of gravity of homogeneous 
bodies which are bounded by the surfaces of these solids are also 
u ly u 2 , etc., and the ordinate of the centre of gravity of the 
collection of bodies is also u. From art. 65, if w denotes the 
specific weight of the imagined homogeneous bodies, 

(wV 1 +wV 2 + . . .)u=wV 1 'U 1 +wV 2 -u 2 + . . . , 
or 

(V t +V 2 + . . . )U = V 1 U 1 +V 2 U 2 + . . . Q.E.D. 

Proof for surfaces: Let A v A 2 , etc., denote the areas of the 
surfaces, u t , u 2 , etc., the ordinates of their centroids, and u the 
ordinate of the centroid of the group. Imagine now as many 
homogeneous thin plates as there are surfaces, and that one face 
of each plate coincides with one or* the surfaces. Let u/, u 2 , 
etc., denote the ordinates of the centres of gravity of the plates, 
and u' that of the centre of gravity of the collection. If w 
denotes the specific weight of the plates and / their thickness, 
their weights are approximately A x tw, A 2 tw, etc., the approxi- 
mation being closer the smaller t is taken. (Of course, for plane 
plates these expressions are correct for all values of t.) From 
art. 65, 

(A 1 +A 2 + . . .)twu' = A J tw-u 1 '+A 2 tw-u 2 ' + . . ., approximately; 



62 CENTRE OF GRAVITY AND CENTROID. [Chap. III. 

or 

(A ± +A 2 + . . .)u , =A 1 u 1 , +A 2 u 2 +. . . , approximately. 

Now this approximate equality approaches exact equality as t 
approaches zero, i.e., 

lim. [(A 1 +A 2 + . . .)w'] = lim. [A& +A 2 u 2 ' + . . .] 

= lim. [/1^/J + lim. [A 2 u 2 ']+ .... 
or 

(A x +A 2 + . . .)u = A 1 u 1 +A 2 u 2 + . . . q.e.d. 

Proof for lines is very similar to that for areas. 

71. Plane of Zero Moment. — Definition. — A plane with 
respect to which the moment of a volume, area, or length is 
zero will, for convenience, be called a plane of zero moment for 
the corresponding solid, surface, or line. It follows from defini- 
tions that a plane of zero moment for a solid, surface, or line 
contains the centroid of that solid, surface, or line. 

Proposition. — If the form of a solid (surface or line) is such 
that it can be divided into parts which may be paired off in such 
a way that the parts of each pair are equal in volume (area or 
length), and that the lines joining the centroids of the parts of 
each pair are bisected by a plane, then that plane is one of zero 
moment. 

Proof: The moments of the volumes (areas or lengths) of 
the parts constituting a pair with respect to the bisecting plane 
are equal but of opposite sign ; hence the moment of the volume 
(area or length) of the pair is zero, and the moment of the entire 
volume (area or length) is zero. Therefore that plane is one 
of zero moment and it contains the centroid of the solid (surface 
or line). 

Evidently the following are planes of zero moment: 
for a circular arc, any bisecting plane containing a diameter; 
for a sector, any bisecting plane containing a diameter; 
for a triangle, any plane cutting it in a median; 
for a parallelogram, any plane cutting it in a diagonal; 
for a triangular pyramid, any plane containing its vertex and 

median of the base. 




§ III.] CENTROIDS OF SOLIDS, SURFACES, AND LINES. 63 

72. Centroids of Simple Solids and Surfaces. — The centroids 
of such solids and surfaces are determined in the succeeding 
articles. The general method consists in finding enough planes 
of zero moment or lines containing the centroid to locate its 
position. 

73. The Centroid of a Triangle is at the intersection of its 
medians. 

Proof: As before stated, a plane cutting a 
triangle in a median is a plane of zero mo- 
ment. Since the centroid is in such a plane 
and in the plane of the triangle, it is in their 
intersection, i.e., the median line. But there are 
three such median lines, hence the centroid is 
at their intersection. Further, OA', OB' , and 
O.O (fig. 47) equal respectively one-third of AA f , BB', and 
CC , from which it follows that 

the distance of the centroid of a triangle from any side 
equals one-third the altitude measured from that side. 

74. The Centroid of a Parallelogram is at the intersection of 
its diagonals. 

Proof: As before stated, a plane cutting the parallelogram 
in a diagonal is a plane of zero moment. Since the centroid is in 
such a plane and in the plane of the parallelogram, it is in their 
intersection, the diagonal; hence, etc. 

75. The Centroid of the Surface of a Pyramid is on the axis * 
of the surface at a distance from the base equal to one-third of 
the altitude. 

Proof : Imagine the pyramid cut by numerous planes parallel 
to its base; the part of the surface between any two adjacent 
planes is a frustum of the surface. Conceive the entire surface 
as consisting of elementary frustums; the centroid of any one 
of them and those of the perimeters of its bases approach 
coincidence. But these perimeters are polygons similar to 
the perimeter of the base and the relation between any 
one of them and the intersection of its plane with the 

* By axis of the surface of a pyramid (or cone) is meant the line join- 
ing its apex with the centroid of the perimeter of its base. 



64 CENTRE OF GRAVITY AND CENTROID. [Chap. III. 

axis is precisely similar to that between the perimeter of 
the base and its centroid. Hence the centroid of all the 
polygons and the elementary frustums lie upon the axis, and 
it follows that the centroid of the surface of the pyramid is on 
the axis. 

The centroids of all the faces of the pyramid lie in a plane 
distant one-third of the pyramid's altitude from the base. It 
follows that the plane is one of zero moment, and hence the 
centroid of the surface is in that plane. 

76. The Centroid of the Surface of a Cone is on the axis of 
the surface at a distance from the base equal to one-third the 
altitude. 

Proof : The surface of a cone may be regarded as the limit of 
the surface of a pyramid, the number of whose faces are in- 
creased without limit. It follows that the limiting position of 
the centroid of the pyramid is the centroid of the surface of the 
cone, etc. 

77. The Centroid of a Prism with Parallel Bases is in the 
axis and midway between the bases. 

Proof: Conceive the prism as consisting of elementary 
laminas whose faces are parallel to the bases. The centroid of 
any lamina and the centroids of its faces approach coincidence. 
Obviously, the centroids of the faces are on the axis, hence the 
centroid of each elementary lamina is also, and it follows that 
the centroid of the prism is on the axis. Obviously, a plane 
midway between the bases is one of zero moment, hence the 
centroid is in that plane. 

78. The Centroid of a Pyramid with a Triangular Base is on 
the axis * at a distance from the base equal to one-fourth the 
altitude (see also next art.). 

Proof: As before stated, a plane containing the apex and a 
median line of the base is one of zero moment. But there are 
three such planes, and since the centroid is in each, it is in their 
intersection, the axis. Now each ' ' corner " of the pyramid may 
be in turn considered as a vertex corresponding to which there 



* A line joining the apex of any pyramid or cone with the centroid of 
the base is the axis. 




§ III.] CENTROIDS OF SOLIDS, SURFACES, AND LINES. 65 

is an axis, and the centroid being on each axis, is at their in- 
tersection. 

Two axes, AF and BG, are represented in 
fig. 48. Since EF = EB/s and EG = EA/ 3 , 
GF is parallel to AB and GF = AB/$, and it 
follows that the triangles OFG and OAB are 
similar. Hence OF = OA/s and OF = AF/ A , 
that is, the centroid is one-fourth the 
length of the axis upward from its foot. D 
Also the distance of the centroid from the 
base equals one-fourth the altitude. 

79. The Centroid of Any Pyramid is on the 
axis at a distance from the base equal to one-fourth the altitude. 

Proof : Conceive the entire pyramid as made up of elementary 
frustums; the centroid of any one of them and those of its 
faces approach coincidence. But these faces are surfaces 
similar to the base, and the relation between any one of them and 
its intersection with the axis is precisely similar to the relation 
between the base and its centroid. Hence the centroids of all 
such surfaces and elementary frustums lie upon the axis, and 
it follows that the centroid of the pyramid is in the axis. 

Conceive the base divided into triangles, and the entire 
pyramid as consisting of pyramids with these triangles as bases. 
The centroids of all these component pyramids lie in a plane 
distant one-third of the pyramid's altitude from the base; it 
follows that the plane is one of zero moment for the entire 
pyramid, and hence the centroid is in that plane. 

80. The Centroid of a Cone is in its axis at a distance from 
the base equal to one-fourth the altitude. 

Proof: A cone may be regarded as the limit of a pyramid, 
the number of whose faces are increased without limit. It 
follows that the limiting position of the centroid of the pyramid 
is the centroid of the cone. Hence, etc. 

81. Centroids of Solids and Surfaces Consisting of Simple 
Parts. — The solids and surfaces whose centroids are now to be 
determined consist of parts whose volumes, or areas, and cen- 
troids are known. The solutions are based on the principle of 
moments, art. 70. 



66 



CENTRE OF GRAVITY AND CENTROID. [Chap. III. 



EXAMPLES. 

i. Determine the centroid of a trapezoid whose altitude is 
a and whose minor and major bases are b and B respectively. 

Solution : Imagine the trapezoid divided into two triangles. 
Their areas are Ba/ 2 and ba/2, and the distances from the base 
B to their centroids are a/ 3 and 2(1/3 respectively. The area of 
the trapezoid is (B + b)a/2; and if y denotes the distance be- 
tween its centroid and the base B, 

(B + b)a _ Ba a ba 2 

-y= + a, 

2 2323 



or 



y= 



2b+B 
W+B) 



a. 



The centroid can be determined geometrically in this way: 
Extend in either direction the major base a distance 6, and in 
the opposite direction the minor base a distance B. Then the 
line joining the ends of the extensions intersects the line joining 
the centres of the bases at the centroid of the trapezoid. The 
student should supply a proof. 

2. Determine the centroid of the "tee section" represented 
in fig. 49. Ans. 1.01 in. above base. 

3. Determine the centroid of the "channel section" repre 
sented in fig. 49. Ans. 0.79 in. above base. 




4. Determine the centroid of the "angle section" repre- 
sented in fig. 49, with reference to the sides AC and BC. 

5. Locate the centroid of any plane quadrilateral. 
Method: Join the centroids of the two triangles into which 

either diagonal divides it; also join the centroids of the triangles 



§111. J CENTROIDS OF SOLIDS, SURFACES, AND LINES. 67 

into which the other diagonal divides it. ' The intersection of 
those two lines is the centroid sought. Prove. 

82. Centroids of Solids and Surfaces Considered as Parts of 
Other Solids or Surfaces. — The solids and surfaces whose cen- 
troids are now to be determined may be conveniently con- 
sidered as consisting of a simple solid or surface minus one or 
more simple solids or surfaces. It is supposed that the volume 
or area and centroid of each of the simple solids or surfaces is 
known. The principle of moments slightly modified is em- 
ployed thus: Let M t be the moment of the volume or area in 
question, M 2 that of the volume or area of which the first is a 
part, and M 3 that of the remainder; then 

M X = M 2 -M S , 

for, according to the principle of moments, M 2 = M 1 +M 3 . 

EXAMPLES. 

1. Determine the centroid of the shaded part of fig. 50. 

Solution: The shaded area consists of <r __a/ 2 _i_ >< __a/ e ,___^ 
the large square minus the triangle and 
quadrant. Its value is 

a 2 ~a 2 \ a 2 . 

T + - I 6) = r 6 (I2 - 7:) - 

Let the base and left side of the square 

respectively be % and y axes, and x and y 

the coordinates of the centroid. The co- F 1 ^- 50. 

ordinates of the centroid of the triangle and quadrant are 

respectively 

(a/6, 2*2/3) an d (a — 20/3-, 20/37:). (See ex. 4, art. 83.) 
Hence 

a 2 9 a a 2 a ita 2 . 

or 

_ 8-7T 



x=- 



12— 7T 

The student should determine y. 

2. Two circles whose diameters are as 2 to 3 are tangent 
internally. Locate the centroid of the part of the larger circle 
not included in the smaller. 



68 CENTRE OF GRAVITY AND CENTROID. [Chap. IIL 

3. Locate the centroid of a surface bounded by a quarter 
of a circle and the tangents at its extremities. 

Ans. Its distance from either tangent = o.2 2 2r. 

4. Suppose that in fig. 49 three corners of the "angle" are 
rounded as shown, the radii at A' and B' being T 5 B - in. and that 
at C \ in. Redetermine the centroid. 

5. Locate the centroid of a frustum of a cone whose major 
and minor base radii are R and r respectively and whose altitude 

is a - Ans. Distance from larger base is *#£*£*£. 




4 R 2 +Rr+r* 

6. Locate the centroid of the shaded part of fig. 51. 
Solution: Since any plane cutting the 

figure in Ox is a plane of zero moment, the 
centroid is on Ox and y = o. To determine 
x, consider the shaded part as consisting of 
the sector of radius r 2 minus the sector of 
Fig. 51. radius r t , and write the moment equation 

with respect to the yz plane. (See ex. 3, art. 83.) 

Ans. x = --^\ !^-sin— . 

3(V-v)a 2 

7. Locate the centroid of a circular segment. 

Ans. Its distance from the centre of the circle equals 

(base of the segment) 3 / (area of the segment) 12. 

83. Centroids Determined by Integration. — Imagine the solid, 

surface, or line divided into elementary parts. Let ~x, y, z 

denote the coordinates of the centroid of the solid, surface, or 

line. 

For a solid, let V denote its volume and x, y, z the coordi- 
nates of the centroid of any elementary part. Then, according 
to the principle of moments (art. 70), 

Vx = /dV.x, Vy = /dV-y, Vz = /dVz. . . (1) 

For a surface, let A denote its area, and x, y, z the coordi- 
nates of the centroid of any elementary part. According to 
the principle of moments, 

Ax = /dA-x, Ay = ydAy, Az = f&kz. . . (2) 



5 in.] 



CENTROIDS OF SOLIDS, SURFACES, AND LINES. 



69 



For a line, let I denote the length, and %, y, z the coordinates 
of the centroid of any elementary part. According to the prin- 
ciple of moments, 



lx = y , dlx, ly = /dl y, lz = /dlz. 



(3) 



The limits of integration in the above formulas, applied to 
any particular example, must be such that the moments of each 
element of the solid (surface or line) are included in the summa- 
tions. The formulas may be used for determining the centroid 
of a solid (surface or line) provided that the form of it is such 
that the integrations can be performed. They are to be em- 
ployed when the solid (surface or line) cannot be divided into 
parts whose volumes (areas or lengths) and centroids are known. 



EXAMPLES. 

i. Locate the centroid of a circular arc. 

Solution: Evidently the centroid is on OC (fig. 52), i.e., 
y = o. To determined, eq. (3) is used. Since dl = rdd, x = r cos 0, 
and l = ra, 



ra 



•x= I re 

J -a/ 



h 

rdd 



r cos 6, or 



_ 2r . a 

x = — sin — . 
a 2 



For a semicircular arc, a = i8o°, and x = 2r/7z = o.6^yr. 
2. Locate the centroid of a circular arc of 90 . 
Ans. Distances from OA and OB equal 2r/n. 





Fig. 53. 

3. Locate the centroid of a circular sector. 

Solutions: Evidently the centroid is on OC (fig. 53), i.e., 
y = o. To determine x, eq. (2) is used. 
(1) If the elementary area be chosen so that dA=pdddp, the 



7° CENTRE OF GRAVITY AND CENTROID. [Chap. III. 

x coordinate of the centroid of the element is p cos 6 ; and since 

A=ir 2 a, 

Pr r*/* . r a 

hr 2 a-x = I I p 2 cos OdpdO, or x = — sin—. 

Jo J -a/2 3« 2 

(2) If the sector is considered as made up of elementary trian- 
gles, so that dA = %r 2 dd, the x coordinate of the centroid of any 
element is %r cos d. Eq. (2) becomes 

/V2 _ at a 

cos 6 dd, or x — — sin — . 
a/ 2 3« 2 

(3) The student should write the expression for x, choosing the 
elementary area so that dA = dx dy. 

For a semicular area, a = i8o°, and ^ = 4^/3^ = 0.42^ 

4. Locate the centroid of a circular quadrant. 
Ans. Distance from either straight side is 4.r/$7c. 

5. Show by the integration method that the centroid of a 
triangle is one-third the altitude from the base. 

(Suggestion: Take the origin of coordinates at the vertex 
and consider the triangle as consisting of elementary strips 
parallel to the base.) 

6. Locate the centroids of a symmetrical parabolic segment 
whose altitude is a. 

Ans. It is on the axis of the parabola distant fa from the 
vertex. 

7. Locate the centroid of the halves into which the axis 
divides the segment described in the preceding example. 

84. Surfaces of Revolution. — General formula (2) of the pre- 
ceding article is used; it will be advantageous to select the 
y elementary area in a certain way, namely, the 

area described by an elementary part of the 
generating curve. Let the x axis be taken 
coincident with the axis of revolution, fig. 54; 
then the area described by a part of the gen- 
erating curve whose length is ds is 271yds. The 
coordinates x, y, and z of the centroid of this 
Fig <V~" area are evidently x, 0, and o respectively; 
hence 



Ax 



= 27: I yds-x y or % = ~t I ocyds, and y=z=o. 



§111.] CENTROIDS OF SOLIDS, SURFACES, AND LINES. 7 1 

A stands for the area of the surface of revolution, and 

A = 27t I yds. 

The limits of integration must be assigned so that each element 
ds is represented in the integrations. 

EXAMPLES. 

i. Locate the centroid of a segment of a spherical surface. 

Solution: The segment may be considered as generated by 
the revolution of a circular arc, as AC about OC, fig. 52. Since 
x = r cos 6, y = r sin 6, and ds = rdd, 



x = 



7. 



cos 6 sin 6 dd 



I. 



a/ a 



sin 6 dd 



= — I I + cos — . 
2\ 2/ 



2. Show by the integration method that the centroid of the 
surface of a cone is one-third of the altitude from the base. 

85. Solids of Revolution. — General formula (1), art. 83, is 
used; it will be advantageous to select a 
certain elementary volume, namely, that 
generated by an elementary part of the gen- 
erating plane which is included between two 
lines perpendicular to the axis of revolution. 
Thus if, in fig. 55, the generating plane is that 
bounded by the solid curve, and the x axis is 
taken coincident with the axis of revolution, 

dV = 7z(y 2 2 — y t 2 )dx. 
Now the centroid of this elementary volume is 
in the x axis, and its x coordinate is the #in 
the figure; hence 

Vx=7t I (y 2 2 -y 1 2 )dx-x, or %=-y I W-yflxdx. 

V denotes the volume of the solid of revolution, and 

V = 7rf(y 2 *- yi >)dx. 

The limits of integration are to be assigned so that each dV is 
represented in the integrations. 




Fig. 55. 



7 2 



CENTRE OF GRAVITY AND CENTROID. [Chap. III. 



EXAMPLES. 

i . Locate the centroid of a segment of a sphere. 

Solution: The segment may be considered as generated by 
the revolution of one-half of a circular segment, as A CD about 
OC, fig. 56. Since x = r cos 6, dx=—r sin 6 dd. Also y 2 = r sin 
and y 1 = o, therefore, 



x = 



'Jo " 



sin 3 6 cos 6 dd 



/ sm 3 
Jo 



= ir 



sin 4 a/2 



Odd 



2—3 cos a/2 +cos 3 a/2* 



2. Locate the centroid of a paraboloid of revolution whose 
altitude is a. 

Ans. It is in the axis of revolution distant fa from the 
apex. 
.a 





IB 

Fig. 56. Fig. 57. 

3. Locate the centroid of a frustum of a cone. 
Suggestion: The frustum may be considered as generated 
by the revolution of the shaded trapezoid (fig. 57) about Ox. 

aR 2 + 2Rr + $r 2 



Ans. 



x = 



4 R 2 + Rr + r 2 ' 
86. Theorems of Pappus and Guldinus. — I. The area of the 
surface of revolution generated by a plane curve revolved about 
an axis in its plane equals the length of the curve times the 
circumference of the circle described by its centroid. 

Proof: Let A denote the area of the surface, / the length of 
the curve, and y the ordinate of its centroid measured from the 
axis. Then (see fig. 54) 



A 
hence 



= 27i I yd. 



and, from eq. (3), art. S$ t y=T I P^s; 

A = l-27ry (1) 



§ III.] CENTROIDS OF SOLIDS, SURFACES, AND LINES. 73 

II. The volume of a solid of revolution generated by a plane 
figure revolved about an axis in its plane equals the area of the 
figure times the circumference of the circle described by its 
centroid. 

Proof: Let V denote the volume of the solid, A the area of 
the plane figure, and y the ordinate of the centroid of A meas- 
ured from the axis. Then (see fig. 55) 

V = nf(y 2 2 - yi 2 )dx, 
and, from eq. (2), art. 83, 

y = \J (y2-yi)dx'i(y 2 +yi)'y 

hence 

V = A-27iy (2) 

Equations (1) and (2) are available for computing y as well 
as A or V if all other factors in the equations are known. It 
should be remembered that A has different meanings in (1) 
and (2), 

EXAMPLES. 

1. Knowing that the surface of a sphere is 47rr 2 , locate the 
centroid of a semicircular arc. 

2. Knowing that the volume of a sphere is -§7rr 3 , locate the 
centroid of a semicircle. 

3. A circle of radius r is revolved about a line in its plane 
whose distance from the centre is a, a being greater than r. 
Deduce expressions for the surface area and volume of the solid 
generated. 

4. By eqs. (1) and (2), deduce formulas for the surface area 
and volume of a cone. 

5. Show that the area of the spherical segment generated by 
revolving the arc AC, fig. 53, about Ox is 27rr 2 (i —cos a). 



CHAPTER IV. 
ATTRACTION AND STRESS.* 

§ I. Gravitation. 

Every body in the universe attracts every other body, the 
attraction in each case depending upon the masses of the two 
bodies and their distance apart. This attraction is called 
gravity, gravitation, and gravitational attraction. 

87. Law of Gravitation. — Every particle attracts every other 
particle with a force which is proportional to the product of the 
masses of the two particles directly and to their distance apart 
inversely. 

The law may be stated algebraically thus : let F denote the 
attraction, m' and m" the masses, and r the distance ; then 

_ m'm" _ , m'm" 

Fa^ — 5—, or F = k 5-, 

r 2 r 2 ' 

k being a constant whose value depends, as explained below, 
upon the units used to express force, mass, and distance. 

88. Gravitation Constant. — It is shown in art. 97 that two 
homogeneous spheres attract each other as though the mass of 
each were concentrated at its centre, i.e., the attraction is the 
same as that between two particles placed at the centres of the 
spheres, the mass of each being equal to that of the correspond- 
ing sphere. Hence the formula for the attraction between par- 
ticles applies to that between homogeneous spheres, r denoting 

* The principles of mechanics employed in this chapter are mainly 
those relating to composition of forces, and most of the chapter might 
have been distributed as problems on composition in Chap. II. How- 
ever, as it all relates to but two subjects, it is convenient to treat the 
matter together. 

74 



§1.] GRAVITATION. 75 

the distance between their centres ; and the attraction between 
two spheres of unit mass, their centres being unit distance apart, 

is k — — =k. 
i 2 

Therefore, the numerical value of k equals the attraction 
between two homogeneous spheres of unit mass whose centres 
are unit distance apart. 

The numerical value of the gravitation constant, k, hence 
depends upon the units employed for expressing force, mass, 
and distance. If the pound, pound, and foot respectively be 
employed, £ = 3.31 X io" 11 , and for C.G.S. units (see art. ) 

& = 6.65X 10 ~ 8 . It is possible to choose the units of force, mass, 
and length so as to make k = 1 ; two of them may be chosen 
arbitrarily, but such a third unit is an uncommon one, and 
therefore k will be retained in the formula. 

A determination of the gravitation constant involves the 
measurement of the attraction between known masses at known 
distance apart. If F denotes the measured force, m' and m" 
the masses of two homogeneous attracting spheres, and r the 
distance between their centres, then 

k = Fr 2 /m'm" '. 

EXAMPLE. 

In an early determination of the gravitation constant, the 
attracting spheres were of lead, two and twelve inches in diam- 
eter. When their centres were nine inches apart, the attraction 
was measured. What was its value in pounds ? 

89. Density. — By density of a body is meant its mass per 
unit volume. We will denote it by d. 

In a homogeneous body , the mass of a unit volume is the same 
no matter where, in the body, the volume is selected; hence the 
density is constant, and its value is found by dividing the mass 
of the body by its volume, i.e., if m and V denote mass and 
volume of the body respectively, 



7 6 ATTRACTION AND STRESS. [Chap. IV. 

In a heterogeneous body, the density is variable and the ex- 
pression m/V gives the average density; and if Am denotes the 
mass of a portion whose volume is AV, 

Am 

-ttt — average density of the portion. 

If the small portion, as its volume approaches zero, always 
includes a point P, then 

Am 
lim. -jy. = density at P, 

or, if d denotes density at P, 

dm 

a= dv- 

90. Attraction at a Point or "Strength of Field." — By 

attraction at a point due to any body is meant the attraction 
which it would exert upon a particle of unit mass placed at the 
point or upon a homogeneous sphere of unit mass whose centre 
is at the point. This is also known as strength of field, the term 
field being a contraction of "field of force," which means the 
region at all points of which there is an attraction. 

91. Attractions in Some Simple Cases. — To compute the 
attraction at a point it is necessary to compound the gravita- 
tional forces exerted by all the particles of the attracting body 
upon a particle of unit mass placed at the point. This system 
of forces is concurrent, and may be collinear, coplanar or non- 
coplanar; hence to determine the resultant of the system the 
methods of §§ 2, 3, or 6, Chap. II, may be used. 

92. Attraction at a Point on the Produced Axis of a Straight 
Slender Rod. — Let AB (fig. 58) be the rod and P the point. If 

p a q b 

i HP =3 



K — - 1 x ->l 

|«. X- * 



Fig. 58. 

a denotes the area of the cross-section of the rod, and dm the 
mass of an elementary volume such as that represented at Q, 
then dm = dadx. Such an element may be considered as a par- 



§!•] 



GRAVITATION. 



77 



tide, for its dimensions are negligible in comparison with x; 
hence the attraction due to it at P is kdadx/x 2 . Denoting the 
attraction of the whole rod by F, then 

F = kSa r' d 4=k8a(L-L\ =k ™(L-L\ 

J x, X \ X \ x 2/ I \ X l X 2/ 

m and / denoting mass and length of the rod respectively. 

93. Attraction at any Point Due to a Straight Slender Rod. — 
Let AB (fig. 59) be the rod and P the point c 
distant from the rod. If a denotes the cross- 
sectional area of the rod and dm the mass of 
an element such as that represented at Q, 
then dm = dady. Such an element may be 
considered as a particle, since its dimensions 
are negligible compared with r; hence the 
attraction due to it at P is kdady/r 2 . 

The x- and ^-components of this attrac- 
tion are respectively F IG# 59< 



/ 
/ 
/ 

y / , 

y v * 



kdady 



cos 6 and 



kdady 



sin 6. 



It is plain from the figure that r = c/cos 6, and y = c tan 6; hence 
dy = cd6/cos 2 6 and the expressions above can be written 



cos 6 do 



and ■ sin 6 dd. 

c 



If F x and F y denote the x- and ^-components of the attraction 
at P, 



and 



_kda r< 

~ c J-i 
_kda r« 

C J-fL 



a Ja kda. . . 
cos d0 = (sin /9 + sm a ), 



sin 



in kda 
d6 = (cos/? 



cos a), 



If F denote the magnitude of the resultant attraction and cf> its 
angle with the x axis, 



tan <f) = F y /F x = tan 



a-fi 



or (f> = (a—P)/2. 



78 



ATTRACTION AND STRESS. 



[Chap. IV- 



94. Attraction Due to a Circular Ring at a Point on its Axis. 
— Let the ring be that represented in fig. 60 and P the point b 

distant from its centre. Imagine the 
ring resolved into elements as that 
represented at Q, and call the mass of 
one dm. Such an element may be 
treated as a particle, for its dimen- 
sions are negligible compared with PQ. 
The attraction at P due to any ele- 
ment is kdm/(r 2 + b 2 ). It is plain 
from the symmetry that the algebraic 
sums of the y- and ^-components of 
the attractions of all the elements are 
zero, hence the attraction at P due to the ring equals the sum 
of the ^-components. Let <j> denote the angle CPQ, then 
the ^-component of the attraction of an element is 
kdm . 

s^v cos *' 

and if F denotes the resultant attraction, 

k . r . km 




F = 



+ V 



f< 



cos (j> I dm = 



r 2 + V 



cos 



m denoting the mass of the ring. 

95. Attraction Due to a Thin Circular Plate at any Point on 
its Axis. — Let AB (fig. 61) be the plate, 
a its radius, t its thickness, and P the 
point. Imagine the plate as consisting 
of rings, or hollow cylinders, whose 
inner and outer radii are r and r + dr 
respectively. The mass of such a ring 
is 027zrdr-t, and the attraction of the 
ring at P equals 



, d2Tcrdr-t , 

k ^2 , ^2 cos 9 




----. a ; 



Fig. 6] 



(see art. 94). 
then 



Let F denote the attraction of the whole plate, 



F=27tkdt 



£"-, 



COS (f> 



dr. 



§1.] GRAVITATION. 

Since r = b tan <j>, dr = b d(f>/ cos 2 cj>, and 



79 



nkot j ' 



m 



F = 2itkbt I sin <j>d<j> = 2-kot(i — cos a) = 27ik^(i — cos a). 

The expression 271(1— cos <j>) is the value of the solid angle* 
subtended by the plate at P; hence, if co denote that angle, 

F=k A<°- 

For a point very near the plate a is nearly 90 , and 

F = h2Tz-r, approximately. 

96. Attraction Due to a Sph rical Shell at any Point. — Let 
ABC (fig. 62) be the shell, represented by a diametral section, 





/"""~~A~ 










/ / ! \ 


^-^ 






/ a i : \ 








// ! ! \ 

Ae 1 i 1 




c 











) 






w ! ; 


/ 






W ! / 






S\l / 








\ \ y j 








\ ;\ / / 








\>JLy 








>Myn' 






"^ — it 


— -^Sm' 





Fig. 62. 

and P the point. Imagine the shell resolved into elementary 
rings which are cut out from the shell by cones whose common 
apex and axis are and OP respectively. Two of the ring sur- 
faces then are conical and two are zones of the outer and inner sur- 



* The student is reminded that a solid angle is measured by the ratio 
of the area of that part of a sphere, whose centre is at the apex of the 
angle, which is included within the surface bounding the angle to the 
square of the radius of the sphere. Thus if for the solid angle sub- 
tended by the plate at P a sphere of radius R be used, the cone bounding 
the angle cuts from the sphere a segment whose area is 2-R*(i —cos a), 
see ex. 5, art. 86; hence the solid angle equals 27r(i —cos a). 



8o ATTRACTION AND STRESS. [Chap. IV. 

faces of the shell. Let dA denote the area of the inner zone, 
and / the thickness of the shell; then the mass of the ring is 
ddA • t and 

the attraction of the ring at P = y~ cos <£> 

(see art. 94). Let F denote the attraction of the shell, then 



= kStf^ 2 



^dA 



the limits being chosen so that all rings of the shell are included. 
To integrate, we will express dA and cos <f> in terms of r. 
From the figure, 

dA = 2na sin 6 • add = 271a 2 sin 6 dd, 

and r 2 = a 2 + b 2 — 2 ab cos 6, of rdr = ab sin Odd; 

hence 

dA = 2n t-t dr. 
b 

Since a 2 = r 2 + b 2 — 2rb cos <j>, 

cos <f> = (r 2 + b 2 — a 2 )/2br. 
Finally, 

/■ 

The integral equals I dr + (b 2 — a 2 ) I — = . 

The limits of the integration depend on whether the point P 
is within or without the shell. 

(a) When the point is external, 



_ Ttkbat 

F = -p- / - -dr. 



_ nkdat r r 2 + a 
F= 10 



2 h2 



w~i- 



b+a kd^naH km 



b-a 



m denoting the mass of the shell. The final expression shows 
that the attraction at a point outside of a homogeneous shell is 
the same as though its mass were concentrated at its centre. 

When the point is on the surface, b = a and F = k47c-j-, 
A denoting the area of the surface of the shell. 



§11.] ELECTRIC AND MAGNETIC ATTRACTIONS. 81 

(b) When the point is internal, 

Tzkdat rr 2 + a 2 -b 2 l a +t>_ 
b 2 L r Ja-b~°' 

97. Attraction at a Point Due to a Sphere. — If the sphere 
may be resolved into homogeneous shells, the results of the 
preceding article may be employed. 

For exterior points, the attraction due to each shell is the 
same as though its mass were concentrated at the centre and 
hence the attraction due to the sphere is the same as though its 
mass were concentrated at the centre. If m denotes the mass 
of the sphere and b the distance of any exterior point from the 
centre, the attraction at that point is km/b 2 , i.e., the attraction 
varies inversely as the square of the distance from the centre. 

For interior points, the attraction due to each shell which 
includes the point is zero, and the attractions of all shells to 
which the point is external are the same as though their masses 
were concentrated at the centre of the sphere; hence only the 
part of the sphere which is nearer the centre than the point is, 
exerts an attraction at that point, and it is the same as though 
the mass of that part were concentrated at the centre. If m' 
denote the mass of that part and b the distance of the point from 
the centre, the attraction at that point is km'/b 2 . 

If the sphere is homogeneous, m' = d^7rb 3 ; hence the attrac- 
tion at the point is k^xdb, i.e.. it varies directly as the distance 
from the centre. 

§ II. Electric and Magnetic Attractions. 

Electrified bodies attract or repel each other, and so do 
magnetized ones. A study of these forces does not fall within 
the scope of this book, but their laws are so similar to that of 
gravitation that several important propositions relating to 
electric and magnetic attractions can be readily deduced from 
the results derived on gravitation. 

98. Laws of Electro-Static and Magnetic Forces. — (1) Bodies 
similarly electrified or magnetized repel, and those dissimilarly 
electrified or magnetized attract, each other. 



82 ATTRACTION AND STRESS. [Chap. IV. 

This law points out a difference between electric or mag- 
netic force and gravitation; the latter is always attractive. 

(2) The force, attraction or repulsion, between two bodies 
electrified or magnetized is proportional directly to the product 
of the quantities of electricity or magnetism and inversely to 
the square of the distance between them. 

In order that the distance between the bodies may be defi- 
nite their dimensions must be negligible compared with the 
distance between any two points of the bodies.* 

The second law may be stated algebraically; thus, let F 
denote the force, q' and q" the quantities, and r the distance, 
then 

F«*£. or F=klf (I ) 

k being a constant corresponding to the gravitation constant. 
It is customary to employ units of force, quantity, and dis- 
tance which make k equal to unity; then 

' F=3V' <„ 

99. Strength of Field. — By strength of field at any point 
due to any quantity of electricity or magnetism is meant the 
force which that quantity would exert upon a unit quantity 
of electricity or magnetism concentrated at that point. It 
is assumed that the unit quantity would not affect the field. 

100. Analogy between Electrical or Magnetic and Gravita- 
tional Attractions. — Comparing the equation of art. 87 and (1) 
of art. 98, it is seen that they are perfectly similar; hence the 
strength of field due to any distribution of electricity or mag- 
netism may be found by analogy from the gravitational at- 
traction due to a body the distribution of whose mass is simi- 
lar to that of the electricity or magnetism. 

101. Strengths of Field Due to Some Simple Distributions 
of Electricity and Magnetism. — I. Electrified Circular Plate. 



* These statements of the laws are somewhat loose, but they serve the 
present purpose better than the usual forms. The reader is assumed to 
have at least an elementary knowledge of the phenomena of these forces. 



§ II.] ELECTRIC AND MAGNETIC ATTRACTIONS. 83 

— The electricity resides at the surface, and is nearly uni- 
formly distributed, being more "dense" at the edges than 
elsewhere on the surface. If we regard it uniformly distributed, 
the expression for the strength of field at any point on the 
axis of the plate due to the electricity on either face may be 
deduced from the expression for the gravitational attraction 
due to a homogeneous thin plate at any point of its axis, which 

wt 
is k-^-io (art. 95) For strength of field due to the electricity 

A 

on one side, we make k=i, and substitute q for m, q denoting 
the quantity of electricity on the side considered. If F de- 
notes the strength of field, then 

Now q/A is the quantity of electricity per unit area, and is 
called the surface density ; if it be denoted by p, 

F = pco (1) 

For a point at the centre of the face, w = 27z, and 

F=27lp (2) 

If the plate be very thin, so that its thickness is negligible 
compared to the distance between plate and the point of the 
axis considered, the strength due to electricity on both sides 
is called strength due to the plate, and its value is qto/A, q 
denoting quantity on both faces. For such thin plates q/A f 
the quantity of electricity on both sides per unit area of one 
side, is called surface density. With this meaning of q/A, or 
p, equations (1) and (2) may be used to compute strengths 
due to thin plates. 

II. Electrified Sphere. — The electricity resides at the sur- 
face and is uniformly distributed over it; hence the expres- 
sion for the strength of field due to an electrified sphere may 
be deduced from that for the gravitational attraction due to 
a homogeneous spherical shell. 



84 ATTRACTION AND STRESS. [Chap. IV. 

(a) For external points, the expression is km/b 2 (art. 96). 
For strength due to an electrified sphere we make k = 1 , and 
substitute q for m, q denoting the quantity of electricity. Then 
if F denote strength, 

b 2 b 2 A' 

A denoting area of the surface of the sphere and r its radius. 
Now q/A is the surface density; if it be denoted by p, 

F=± VTP • • (3) 

For a point at the surface b = r, and 

F = wp. . (4) 

(b) For internal points, the gravitational attraction due 
to a shell is zero, and hence the strength of field due to an elec- 
trified sphere at an internal point is also zero. 

III. Magnetic Shell. — This is a magnetized plate, the mag- 
netism on one side being ' ' north-seeking ' ' and that on the 
other side "south-seeking." The expression for strength of 
field due to the magnetism on one side of a circular plate at a 
point on its axis may be deduced from the expression for the 
gravitational attraction due to a homogeneous thin circular plate. 

fyi 
That expression is k-rco (art. 95). For strength of field due 
A 

to the magnetism on one side we make k = 1 , and substitute 

q for m, q denoting the quantity of magnetism. Then if F 

denote strength, 

*- !<, 

Now q/A is the quantity of magnetism on one side per unit 
area, and is called the surface density of magnetism; if it is 
denoted by p, 

F=p(o (5) 

For a point at the surface of the shell co = 27t t and 

F=27Zp (6) 



§ III.] STRESS. 85 

§ III. Stress. 

102. Stress Defined. — The term stress is variously defined. 
Some writers mean by it the forces which any two bodies or 
two parts of a body exert upon each other, it being then a 
term which refers to any "action" and its "reaction" (art. 6). 
For example, such writers designate as a stress the forces which 
the earth and sun exert upon each other, the forces which the 
upper and lower halves of a monument exert upon each other, 
etc. 

Most engineers, however, use the term in a narrower sense, 
meaning by it the force which one part of a body exerts on an 
adjacent part at the surface of contact of the parts. Such 
engineers designate as stresses the force which the upper or 
lower half of a monument exerts upon the other half, the force 
which either half of a stretched string exerts on the other half, 
etc. 

We will use the term in the engineer's sense slightly ex- 
tended and define it thus: Stress is any force whose place of 
application is a surface. The force may be exerted between 
parts of one body or between two different bodies which are 
in contact, a part or all of the contact surface being the place 
of application of the force. 

103. Units for Expressing Stress. — Since a stress is a force, 
it must be expressed in force units, the pound force, the kilo- 
gram force, etc. 

104. Classification of Stresses. — If the parts of a stress on 
all equal small portions of its place of application are equal, 
the stress is said to be uniform as to distribution; if otherwise, 
non-uniform. If the action lines of the resultant forces on 
all the small portions are parallel, the stress is said to be uni- 
form as to direction; if otherwise, non-uniform. 

If the action lines of the resultants are all normal or tan- 
gential to the surface of application , the stress is called simple; 
if otherwise, complex. Stresses are, in general, complex, but 
it is possible to describe any complex stress in terms of simple 
stresses; only such are discussed herein. 

Simple stresses may be classified into normal and tangen- 



S6 ATTRACTION AND STRESS. [Chap. IV. 

tial stresses according as the action lines of all the forces 
on the small portions of the place of application are normal 
or tangential to the surface to which the stress is applied. 
Normal stresses are subdivided into pressures, or compressions, 
and tensions according as the force, or stress, acts toward or 
away from the place of application. A tangential stress is 
also called a shear. 

The classification may be presented thus: 

f -, ( pressure 

v normal ...if 
Simple stresses < ' tension 

( tangential . . shear 

105. Description of a Simple Stress. — This requires a state- 
ment as to its kind (pressure, tension, or shear), and as to the 
manner of its distribution. The distribution is described by 
a statement, for each point of the place of application, of the 
value of the 

106. Intensity of Stress. — By intensity of stress, or stress 
intensity, at any point of the place of application of the stress 
is meant the stress per unit area at that point. 

If the stress is uniform as to distribution, then the stress. 
per unit area is the same at all points, and its value is found 
by dividing the stress by the area of its place of application. 
If F, A, and p denote the stress, area, and intensity respect- 
ively, 

P = F/A (1) 

If the stress is non-uniform as to distribution, then the 
stress per unit area is different at different points, and the 
expression F/A gives the value of the average intensity. Also 
if A A denotes the area of any part of the place of application 
and AF is the value of the stress applied to that part, AF/ AA 
is the value of the average intensity of that part, AF, of the 
whole stress. Now if A A, as it approaches zero, always in- 
cludes a point P, then AF/ AA approaches, in general, a finite 
limit, and the value of that limit is the intensity of stress at 
P, or 

dF 
P = dA (2) 



§ni.] 



STRESS. 



87 



The unit of intensity of stress depends on the unit used for 
expressing F and A. If, for example, the pound and square 
inch are used for these respectively, then the pound per square 
inch is the corresponding unit for stress intensity. 

[Note: What is herein called intensity of stress is very 
commonly called by engineers "unit stress." Strictly, unit 
stress is the general name for the units employed for express- 
ing stresses, the pound, kilogram, etc., and the engineer's 
usage is avoided in this book as being confusing to the student. 
Once thoroughly familiar with the quantities involved, he may 
safely adopt the engineer's term.] 

107. Graphical Representation of a Simple Stress.* — 

(a) Normal Stress. — Let abed, in the xy plane (fig. 63), be 
the place of application of the stress, and imagine ordinates 
erected at all points of it propor- 
tional to the intensities of stress at 
the points; thus, if p denote the in- 
tensity at P, z the ordinate, and k 
any constant (the scale number), 

p = kz, or z = p/k. 
Then the volume of the solid defined 
by those ordinates represents the 
stress, for its altitude at any point represents the intensity 
there, and the volume represents the value of stress, as can 
be shown, thus: 

the stress, or 




F = fpdA=kfzdA, 
and the volume, or V= I zdA ; 



hence 



F = kV, or V=F/k. 



If the stress is partly tensile and partly pressural, the ordi- 
nates corresponding to tensions and pressures are drawn from 
the plane in opposite directions. Then p is regarded as posi- 



* In this and the following articles it is assumed that the place of 
application of the stress is plane, but some of the results are not restricted 
to such cases. 



88 



ATTRACTION AND STRESS. 



[Chap. IV. 



tive for one kind and negative for the other kind of stress. 
(See fig. 64.) 

(b) Tangential Stress. — A tangential stress which is uni- 
form as to direction may also be represented by the method 
described above. 

108. Centre of Stress. — Any stress which is uniform as to 
direction can be conceived as a system of parallel forces having 
definite application points. Thus imagine the place of applica- 
tion of such a stress divided into elementary parts, and let dA 
denote the area of any element and p the intensity of stress at 
that element; then the stress or force on that element is pdA. 
Now all such forces as pdA make up a system of parallel forces 
and their application points are definite points of the place 
of application of the stress. 

Since a system of parallel forces having definite applica- 
tion points has a centroid (art. 62), therefore a stress which is 
uniform as to direction has a centroid, or a centre as it is more 
commonly called. 

The centre of stress is in the plane of the place of applica- 
tion, and it is where the action line of the resultant of the ele- 
mentary forces, pdA, pierces that plane. Formulas for the 

position of the centre of stress in the 
plane may be deduced from those for 
the centroid of a system of parallel 
forces (art. 63). Thus, let fig. 64 rep- 
resent a stress and x c and y c the coor- 
dinates of the centre of the stress. 
Also, let x and y denote the coor- 
dinates of the application point of the force pdA on any 
elementary area dA . Then 




Fig. 64. 



*<; = 



/pdA 



and 



/pdA-y 



F JC F ' 

F denoting the value of the stress, or I pdA. 

Proposition. — A line drawn through the centroid of the 
solid representing a stress and perpendicular to its place of 
application passes through the centre of the stress. 



I in.) 



STRESS. 



89 



Proof: According to the preceding article, the equations 
above can be rewritten thus: 



Xr = 



fzdA-x fzdA-y 
rr — and y c = 



V 



hence 



fdV-x fdV-y 

*c= — y— and y c = — y — . 

fdV-x fdV-y 

Now — p — and — ^ — are the expressions for the coordi- 
nates of the centroid of the solid whose volume is V (art. 83); 
hence the corresponding coordinates (x and y) of the centre 
of stress and the centroid of the solid representing the stress 
are equal. It follows that the line joining those two points 
is parallel to the z axis, hence, etc. 

109. A Uniformly Varying Normal Stress. — By this is 
meant one whose intensity at any point is proportional to the 
distance of that point from some straight line in the plane of 
the place of application of the stress. The straight line is 
called a neutral axis, or zero line. A familiar example of such 
a stress is the pressure of a liquid which is at rest upon an im- 
mersed flat surface which is not horizontal, and an important 
case is the "fibre stress" on any cross-section of a moderately 
loaded beam. 




Fig. 65 represents three such stresses; the places of appli- 
cation are abed, and the zero lines are coincident with the y 
axes. In (a) the stress is part tension, part pressure; in (b) 



9° ATTRACTION AND STRESS. [Chap. IV. 

and (c) it is all of one kind. In any case, the law of variation 
can be expressed thus: 

p=ax, 

p being the intensity at a point whose distance from the zero 
line is x and a sl constant. When the stress consists of a ten- 
sion and a pressure, then p in the equation must be regarded 
as having sign, its sign being the same as that of x. 

When x = i , p = a; hence the numerical value of a equals 
the value of p at points at unit distance from the zero line. 

The value of the stress. — The stress on an elementary area 
is 

dF=pdA=axdA, 
hence 

F = afxdA (i) 

Since fxdA=xA, A being the area of the place of application 

of the stress and x the x coordinate of its centroid, 

F = axA (2) 

Let p denote the intensity of stress at the centroid of the place 
of application, then 

p = ax and F = pA (3) 

Proposition. — The intensity of a uniformly varying stress 
at the centroid of its place of application and the average in- 
tensity of the stress are equal. 

Proof: From equation (3), p = F/A; since F/A is the value 
of the average intensity (art. 106), the proposition is proved 
by that equation. 

no. Centre of a Uniformly Varying Stress. — Substituting 
values of p and F for a uniformly varying stress in the first 
expressions for the x c and y c in art. 108, we have 

af(xdA)x fx 2 dA 
aAx Ax 



x c = 



af(xdA)y ixydA 
y ° = aAx Ax 



§111.] STRESS. 91 

Ax is the moment of the area of the place of application of the 
stress with respect to the zero line. The integrals in the ex- 
pressions for x c and y c are of great importance and have special 
names; they are discussed in App. D, where their values for 
many forms of surfaces are deduced. 

EXAMPLES. 

i. Where is the centre of a stress whose place of applica- 
tion is a rectangle, the intensity at any point being propor- 
tional to its distance from one side? 

2. Where is the centre of a stress whose place of applica- 
tion is a semicircle, the intensity at any point being propor- 
tional to its distance from the diameter ? 

in. Moment of a Uniformly Varying Normal Stress. — The 
stress on an element of area dA whose ordinate is x, is axdA, 
and its moments with respect to the x and y axes are respect- 
ively 

(axdA)y and (axdA)yx. 

If M x and M y denote the moments of the entire stress about 
the x and y axes respectively, 

M x = a fxydA 
and 

M y = ajx 2 dA. 

112. Position of the Neutral Axis when the Stress is Part 
Tensile and Part Pressural and the Two Parts are Equal. — The 
algebraic sum of the elementary stresses is zero; hence 

axA=o, or ~x = o } 

i.e., the zero line contains the centroid of the surface of appli- 
cation of the stress. 

In a beam, as ordinarily loaded, there is on each cross- 
section a stress of the kind described in the title to this article, 
and the result above deduced is of great practical importance. 
It will bear another deduction; it differs but slightly from that 
given above. Let A' and A"_ be the areas of the parts of the 



92 ATTRACTION AND STRESS. [Chap. IV. 

place of application of the stress which sustains tension and 
pressure respectively, and ~% r and ~x" the distances (not coor- 
dinates) of their centroids from the zero line. Then 

the value of the tension is ax'A'\ 

" " pressure is ax" A" (art. 109). 

Since these are equal, x'A' = x"A", that is, the moments of 
the two parts of the entire area A about the yz plane are equal; 
hence that plane contains the centroid of the whole area (art. 
71). But the centroid is also in the xy plane, hence it is in 
the y axis, or the neutral axis. 



CHAPTER V. 
GENERAL PRINCIPLES OF EQUILIBRIUM. 

§ I. Preliminary. 

113. Definitions. — An external force is one exerted on a 
body by some other body. An internal force is one exerted 
on a part of a body by some other part. The same force may 
be classed either as external or internal, depending upon the 
point of view. For example, consider a block resting upon a 
table; the two bodies exert forces, or pressures upon each 
other. If the block and table are considered as one body, then 
both of these forces are internal, but if they are considered as 
two separate bodies, then each of the forces is an external one. 
Again, the upper and lower halves of the block exert forces 
upon each other, and with reference to the whole block con- 
sidered as one body each is an internal force; but with refer- 
ence to either half considered as a body, the force exerted 
upon that half is an external one. 

A body is said to be in equilibrium if (1) it is at rest or (2) 
its state of motion is unchanging. The first is the important 
case in this connection. 

The system of external forces applied to a body which is in 
equilibrium is also said to be in equilibrium. 

114. General Condition of Equilibrium for a System of Forces 
Applied to a Rigid Body. — If the system of external forces 
applied to a rigid body is in equilibrium, its resultant is nil. 
For the system, being in equilibrium, produces no change of 
motion, and therefore its resultant would produce none; hence 
the resultant must be nil. 

Conversely, if the resultant of all the external forces applied 
to a rigid body is nil, that system of forces is in equilibrium. 

93 



94 GENERAL PRINCIPLES OF EQUILIBRIUM. [Chap. V. 

For the resultant, being nil, would produce no motion, there- 
fore its equivalent, the system, produces none; hence it is in 
equilibrium. 

The condition fulfilled by a system in equilibrium and the 
condition to insure the equilibrium of a system are one and 
the same, namely, the resultant is nil; this is called, therefore, 
the general condition of equilibrium. 

115. Equilibrium of a System of Forces Applied to a Non-rigid 
Body. — The condition stated in the preceding article for the 
equilibrium of forces applied to a rigid body is, as shown, both 
necessary and sufficient. For the equilibrium of a deform- 
able body it is necessary that the resultant of the external 
applied forces be nil, but it is not sufficient. This may be 
explained by illustration: 

Consider the water in a cup; the system of external forces 
applied to it consists of its weight, the pressures exerted by 
the cup, and the air pressure on top. Now if this same sys- 
tem of forces could be applied to the water when frozen, it 
would certainly be in equilibrium and its resultant would be 
nil; hence the resultant of the external system on the body of 
water is nil. But any forces applied to the body of ice which 
with the weight have a zero resultant will maintain its equi- 
librium; thus, a single vertical force equal to the weight and 
acting through the centre of gravity will answer. The same 
force applied to the body of water will not of course maintain 
its equilibrium, although the resultant of the two applied 
forces is nil. Other conditions than a vanishing resultant 
therefore are necessary. In this case, pressure must be ex- 
erted on the entire surface of the body of water except the top, 
and in a certain manner. 

Again, consider two similar boards fastened together with 
one nail as in fig. 66(a) and two fastened with several nails 
as in fig. 66(6), and suppose their planes vertical. It can be 
proved experimentally and otherwise that the equilibrium 
of the first two boards can be maintained by two forces such 
as R r and R" . Clearly, the same two forces would maintain 
the equilibrium of the second two boards ; hence the resultant of 
the system R f , R" , W, and W" is zero. But any forces which 



§11.] CONDITIONS OF EQUILIBRIUM. 95 

together with W and W" have a zero resultant would main- 
tain the equilibrium of the boards in fig. 66(6); thus, a single 
vertical force, as R, equal to W f and W" will answer. The same 
force applied to the first two boards will not of course main- 
tain their equilibrium, although the resultant of the applied 





R 



P>) 



(a) 

Fig. 66. 
forces is zero. Therefore, other conditions than a vanishing 
resultant are necessary. 

Summing up, if the system of external forces applied to a 
deformable body is in equilibrium, the resultant is nil; but 
the converse is not necessarily true. 

§ II. The Conditions of Equilibrium for the Various 

Classes of Force Systems. 
N 116. Collinear Forces. — The algebraic condition of equilib- 
rium is, 2"F = o, 
i.e., the algebraic sum of the forces equals zero. 

The graphical condition of equilibrium is that the force 
polygon for the forces closes, or, the vector sum of the forces 
equals zero. 

For if IF equals zero or the force polygon for the forces 
closes, the resultant is nil. 

117. Coplanar Concurrent Non-Parallel Forces. — The algebraic 
conditions of equilibrium may be expressed in various ways : 
(<*) 2^ = 0, I¥ y = o; (1) 

i.e., the algebraic sum of the resolved parts of the forces 

along two lines, x and y, equals zero. 
(b) 21^ = 0, IM = o; (2) 

i.e., the algebraic sum of the resolved parts of the forces 

along a line x equals zero and the moment sum for the 

forces with respect to a point equals zero. (The direction 



9$ GENERAL PRINCIPLES OF EQUILIBRIUM. [Chap. V. 

x is not to be perpendicular to the line joining the common 
intersection of the action lines of the forces with the origin 
of moments.) 
(c) IM a =o, 2*M 6 = o; (3) 

i.e., the moment sums of the forces for each of two origins 
equal zero. (The origins and the common intersection of 
the action lines of the forces are not to be collinear.) 
For, in either case the resultant is zero, as may be thus ex- 
plained : 

(a) According to art. 36, the resultant R of the system, if 
there is one, is given by 

R = (IF x + IF y ) *; hence, if IF X and IF y equal zero, R 
equals zero. 

(b) If IM a is zero, then the resultant, if there is one, must 
pass through a as well as through the common point 
of the action lines of the forces, 0. If the angle between 
Oa and the x axis be a, R X = R cos a = IF x \ and since 
IF x = o and a is not go°, R must be zero. 

(c) As before, the resultant, if there is one, must pass through 
and a; but if 2M h is also zero, and 0, a, and b are not 
collinear, R must be zero. 

The graphical condition of equilibrium is that the force 
polygon for the forces closes, i.e., the vector sum of the forces 
equals zero. For, if the polygon closes, the resultant is zero, 
see art. ^^. 

118. Special Condition of Equilibrium for Three Forces. — The 
following form of the algebraic conditions 
is often more convenient of application 
than the conditions given in the pre- 
ceding article. 

F^sin a x = F 2 /sin a 2 = F 3 /sin a 3 ; 

i.e., each force is proportional to the sine 

of the angle between the other two (see 

fig. 67). For, from the force polygon for 

Fig. 67. the f orces> w hich closes (fig. 676), 

FJsm (iSo-a 1 )=F 2 /sm (i8o-a 2 ) = F 3 /sin (i8o-a 3 ), or 
FJsin a 1 = F 2 /sin a 2 = iVsin a 3- 




§11] CONDITIONS OF EQUILIBRIUM. 97 

119. Coplanar Non-Concurrent Parallel Forces. — The alge- 
braic conditions of equilibrium may be expressed in several ways : 

(a) 2T = o, 2M = o; (1) 

i.e., the sum of the forces and the sum of the moments 
of the forces each equal zero. 

(b) 2 , M a = o, 2M 6 = o; (2) 

i.e., the moment sums for the system for each of two 
origins equal zero. (The line joining the origins is not to 
be parallel to the forces.) 
For in either case the resultant is zero, as may be thus ex- 
plained: In art. 42, it is shown that the resultant of such a 
system is either a force or a couple, and if a couple, its moment 
equals the algebraic sum of the moments of the given forces 
about any point. 

(a) If IF equals zero, the resultant is not a force, and 
if IM equals zero, the resultant is not a couple, hence the 
resultant vanishes. 

(b) If 2M a equals zero, the resultant is not a couple and 
the resultant force, R, (if there is one,) must pass through 
a. But if IMi equals zero, R must equal zero, or pass 
through 6; but R is parallel to the forces of the system 
and cannot therefore pass through a and b. Hence R 
equals zero, i.e., the resultant vanishes 

The graphical conditions of equilibrium are (1) the force 
polygon and (2) the funicular polygon for the forces must dose. 
For, if the force polygon closes, the resultant is not a force, 
and if the funicular polygon closes, it is not a couple (see art. 40) , 
hence the resultant vanishes. 

120. Coplanar Non-Concurrent Non-Parallel Forces. — The 
algebraic conditions of equilibrium are 

(a) i , F x = o, I¥ y = o f .01 = 0; . . . . (t) 

i.e., the algebraic sums of the resolved parts of all the 
forces along each of two lines and the moment sum of the 
forces with respect to any origin equal zero. 



98 GENERAL PRINCIPLES OF EQUILIBRIUM. [Chap. V. 

(b) i'F^o, ZM a = o, IM b = o; .... (2) 

i.e., the algebraic sum of the resolved parts of all the 
forces along any line and the moment sums for the system 
with respect to two origins equal zero. (The direction of 
resolution is not to be perpendicular to the line joining 
the two moment origins.) 

(c) 2M a = o, IM b = o, IM c = o; ... (3) 

i.e., the moment sums for the system with respect to 
three origins equal zero. (The three moment origins are 
not to be collinear.) 

For, in either case the resultant is nil, which may be proved 

as follows: It is shown in art. 47 that the resultant of a system 

of this kind is a force or a couple; if a force, its magnitude 

2 ■ 2 1 

equals {IF X + 2F y p, and if a couple, its moment equals 

2M. 

(a) If IF X and IF y equal zero, the resultant is not a force, 
and if IM equals zero, the resultant is not a couple; hence 
the resultant vanishes. 

(b) If ZM = o, the resultant is not a couple, and the result- 
ant force, R, (if there is one) must pass through a and 
b in order to make IM a and IM^ equal to zero. If now 
ZF X (=R X ) equals zero, R must equal zero. 

(c) If ZM = o, the resultant if there is one, is not a couple, 
but a force. If IM a , IM h , and IM C equal zero, this re- 
sultant force must equal zero, or else pass through a, 6, 
and c. Since the latter case is impossible, the resultant 
vanishes. 

The graphical conditions of equilibrium are (1) the force 
polygon and (2) the funicular polygon must close. For, if the 
force polygon closes, the resultant is not a force, and if the 
funicular polygon closes, it is not a couple; hence the resultant 
vanishes (see art. 45). 

121. Condition of Equilibrium for Four Forces. — The re- 
sultant of either pair balances that of the other pair. This 



§11.] CONDITIONS OF EQUILIBRIUM. 99 

special condition may often be advantageously employed in 
graphical solutions of problems. 

122. Non-Coplanar Concurrent Forces. — The algebraic con- 
ditions of equilibrium are: 

IF x = o, I¥ y = o, 2T 2 = o; 

i.e., the algebraic sums of the resolved parts of all the forces 
along three directions, not coplanar, equal zero. For, in art. 50 

it is shown that the resultant R of this kind of a system is given 

2 — 2 2 1 

by R = (IF Z +IF y + IF Z )*; hence if the conditions be ful- 
filled, R must equal zero. 

The graphical conditions of equilibrium are that the force 
polygons for two projections of the vectors representing the 
given forces, regarded as force systems, must close. For, 
the closing lines of such polygons are the projections of the 
resultant of the given forces, and if the force polygons 
close, the projections vanish and the resultant is zero (sec 
art. 49). 

123. Non-Coplanar Non-Concurrent Parallel Forces. — The 
algebraic conditions of equilibrium are : 

IF = o y i , M x = o, IM y = o; 

i.e., the algebraic sum of the forces and their moment sums 
with respect to two axes equal zero. (Neither of the axes 
is to be parallel to the forces.) For, in art. 54 it is shown 
that the resultant, if there is one, is a force or a couple. If 
IF is zero, the resultant is not a force, and. if 2M X and 2M y 
equal zero, the resultant is not a couple; hence the resultant 
vanishes. 

The graphical conditions of equilibrium are (1) the force 
polygon for the forces and (2) the funicular polygons for the 
projections of the vectors representing the given forces, regarded 
as forces, on two planes must close. For, the resultant of the 
system, if x there is one, is a force or a couple; if the force poly- 
gon closes, the resultant is not a force, and if the funicular 
polygons close , it is not a couple ; hence the resultant vanishes 
(see art. 52). 






ioo GENERAL PRINCIPLES OF EQUILIBRIUM. [Chap. V. 

124. Non-Coplanar Non-Concurrent Non-Parallel Forces. — 

The algebraic conditions of equilibrium are: 
IF x = o, I¥ y = o y 2F 2 = o, 21^ = 0, 2M 2/ = o, IM 2 = o; 

i.e., the algebraic sums of the resolved parts of the forces 
along three lines and the moment sums for the system with 
respect to three axes equal zero. (The three lines must not be 
coplanar, nor the three axes.) 

For, in art. 55 it is shown that the resultant of such a system 
is a force and a couple, and in art. 58 that the magnitude of 
the force is (IF* + IF J + JF, 2 )^ , and the moment of the 
couple is (lM~x + I My 2 + IM e 2 )* ; hence, if the conditions are 
fulfilled, the resultant vanishes. 

The graphical conditions may be expressed thus: If the 
system be resolved into two component systems, one coplanar 
and one parallel, the forces of the latter being perpendicular 
to the plane of the former, then the following polygons must 
cl~:3: 

the force polygon for the coplanar system, 
1 ■ funicular polygon for the coplanar system, 
" force polygon for the non-coplanar system, 
" funicular polygons for the projections of the vec- 
tors representing the parallel system on each of 
two non-parallel planes. 
For, if these conditions are fulfilled, the resultants of the compo- 
nent, systems are zero (see arts. 45 and 52), and hence the re- 
sultant of , the system itself vanishes. 

125. Special Condition of Equilibrium and Summary. — 
Proposition.- — If three forces are in equilibrium, they must 

be coplanar and concurrent or parallel. 

Proof: The resultant of the first two forces must be a single 
force, since it must balance a force, the third one. Since the 
two forces and their resultant are coplanar and the resultant 
and the third force are collinear, the three given forces are co- 
planar. If the first two forces are parallel, their resultant is 
parallel to . them, and hence the third force is also; if the first 
two forces, intersect, their resultant passes through their inter- 
section, and hence the third force does also. 



§ II.] CONDITIONS OF EQUILIBRIUM. 101 

Algebraic Conditions of Equilibrium. 

Coplanar Systems : 

Collinear IF = o. 

Concurrent Non-parallel IF x = IF y = o, 

IF X =IM = o, or 

IM a = IM b = o. 
Non-concurrent Parallel IF =IM = o, or 

IM a =IM b =o. 
Non-concurrent Non-parallel IF x = IF y = IM = o, 

IF X = IM a = IM b = o, or 

IM a = IM b = IM c = o. 
Non-coplanar Systems: 

Concurrent IF x = IF \ = IF Z = o. 

Non-concurrent Parallel IF = IM x = IM y = o. 

Non- concurrent Non-parallel . IF x = IF y = IF Z = o and 

IM x = IM y = IM g = o. 

Graphical Conditions of Equilibrium. 
Coplanar Systems : 

Concurrent The force polygon closes. j 

Non-concurrent.. .The force and funicular polygons close. 

Non-coplanar Systems : 

Concurrent The polygons for the projections of the 

vectors representing the system on any 
two non-parallel planes close. 

Non-concurrent 

Parallel The force polygon closes and the funicular 

polygons for the projections of the. vec- 
tors representing the system on two non- 
parallel planes close. 

Non- concurrent • 

Non-parallel. . ..If the system be resolved into component 
systems, one coplanar and one parallel, 
the forces of the latter being normal, to 
the plane of the former, then the appro- 
priate conditions stated above apply to 
each component system. 



CHAPTER VI. 
APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. 

§ I. Preliminary. 

126. Nature of the Problems. — The problems involved in 
the application of the principles of equilibrium are usually of 
this kind: a system of forces is in equilibrium and some of 
them are partly or wholly unknown ; it is required to determine 
the unknown elements. The required elements may be the 
magnitudes, directions, or action lines of forces. 

127. General Method of Solution. — These problems can be 
solved by two methods, the algebraic and the graphical. The 
algebraic method is to write the appropriate equations of equi- 
librium for the kind of a force system under consideration, and 
then to solve them for the unknown quantities. This process 
is called applying the algebraic conditions of equilibrium. The 
graphical method is to apply the appropriate graphical condi- 
tions of equilibrium for the kind of a force system under con- 
sideration. How these conditions are applied is explained in 
the solution of some of the following examples. 

Often the force system in equilibrium of which the partly 
or wholly unknown forces are a part, and to which the condi- 
tions of equilibrium are to be applied, is not specified. In such 
cases the system may be recognized by directing one's attention 
to a body which is in equilibrium and with reference to which 
some or all the unknown forces are external. All the external 
forces applied to that body constitute a system in equilibrium. 

In all but the simplest of the following examples the student 
is strongly urged to make a sketch representing the body con- 
sidered and the external forces exerted upon it before apply- 
ing the conditions of equilibrium. He will be aided in his 

102 



§ II.] FLEXIBLE CORDS. 103 

enumeration of the external forces if he will represent first the 
actions through distance * exerted upon the body by other 
bodies and then count the number of contacts between the 
body under consideration and other bodies; at each place of 
contact a force may be exerted upon the body (art. 7). 

§ II. Flexible Cords. 

128. Definitions. — A perfectly flexible cord is one which 
may be bent without resistance. Such a cord is ideal, but 
some cords are practically perfectly flexible; such, for brevity, 
will be called flexible, other cords being called stiff. A heavy 
flexible cord if unsupported cannot be stretched straight, but 
will sag more or less, depending upon the applied pulls. The 
lighter the cord, the less the sag; and, if the weight of the cord 
be small compared with the pulls, it will be practically straight. 
In the following, cords will be assumed without weight except 
where otherwise stated. 

129. Tension in a Cord. — The phrase tension in a cord refers 
to the forces which two parts of a taut cord exert upon each 
other. The forces are equal and opposite (see art. 6). By 
magnitude of the tension is meant the magnitude of either force. 

To illustrate, suppose that AB (fig. 68) is a flexible cord 
without weight subjected to equal a c b 

pulls at its ends, and imagine a plane p' p" 

of separation at any place, C, between Fig. 68. 

the ends of the cord. Since the part AC is in equilibrium, a 
force acts upon it at its right end equal and opposite to P'\ 
this force is exerted by the part BC. Similarly, a force acts on 
BC at its left end equal and opposite to P" ; this force is ex- 
erted by the part AC. These two equal and opposite forces at 
C hold the parts AC and BC together. The magnitude of the 
tension is P' (or P") no matter where C is taken. 

If the pulls P' and P" are unequal, the cord will not be in 
equilibrium, and the magnitude of the tension varies with 
the section C . 

* The only actions through distance considered in this chapter are 
the weights of bodies. 



104 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



EXAMPLES. 

i. A body weighing ioo lbs. is suspended by a single cord 

which is deflected from the vertical by a horizontal force of 

20 lbs. applied to the body. How great are the deflection and 

the tension in the cord? 

Solutions: (1) Algebraic. Considering the forces acting 

upon the body, it is seen that they are three in number, namely, 

• its weight, the horizontal force, and the pull of 

J the string. The conditions of equilibrium, with 

T V'j axes as in fig. 69, are (art. 125) 

«L__\L|? lbs - IF x = 2o-Tsmd = o; 

I IF y = —100 + T cos d = o. 

100 lbs. Solving these it is found that 

Fig. 69. 

T= 102 lbs. and #=ii°io'. 

(2) Graphical. The condition of equilibrium is that the force 
polygon for the three forces must close (art. 125). To construct 
the polygon, the wholly known forces are represented first; 
thus AB and BC (fig. 70) represent the weight and the hori- 
zontal pull respectively, scale 1 in. = 100 lbs. 
The closing side CA then represents the 
magnitude and direction of the pull of the 
string exerted upon the body. 

2. In the preceding example, how great 
a horizontal force would be required to 
deflect the cord 30 , and how great would 
the corresponding tension be? 

3. The two ends of a cord are fastened to 
hooks in the same horizontal line, and at 
the middle a second cord is knotted which sustains a freely 
hanging body weighing 100 lbs. The distance between the 
hooks being such that the first cord makes angles of 20 with 
the horizontal, determine the tension in each half of the first 
cord. Ans. 146 lbs. 

4. Call the angle in the preceding example a and the weight 
of the body W. Deduce an expression for the tension. 




§11] FLEXIBLE CORDS. 105 

5. Suppose that the second cord in ex. 3 is knotted to the 
first at such a point that the inclinations are 10 and 50 degrees. 
Determine the tensions, employing the special condition of 
equilibrium of art. 118. Ans. 74.2 lbs. t 

6. Fig. 71 represents an upright frame within which is 
arranged a network of cords; the connec- A B 
tion EC consists of a spring balance. 
Suppose that the balance, wrong end up, 
reads 10 lbs. Determine the tension in 
AF. Ans. 26Jlbs. 




D C 

7. A given bar is to be supported in p IG 
a certain position by two strings fastened 

to its ends. What condition must the directions of the strings 
satisfy? (See prop., art. 125.) 

8. A uniform bar weighing no lbs. is supported in a hori- 
zontal position by strings fastened to its ends. If one is in- 
clined 30 to the bar, what is the inclination of the other and 
what are the tensions ? 

9. A uniform plate triangular in shape is suspended in a 
horizontal position by vertical cords, one being fastened to 
each corner. Show that each cord sustains one third the 
weight of the board. 

10. Three hooks, A, B, and C, in a ceiling lie on a circle of 
3 ft. radius, and the arcs A B, BC, and CD are equal. A small 
ring is suspended at a point 4 ft. below the centre of the circle 
by means of three cords fastened to the hooks. If a body 
weighing 100 lbs. is suspended from the ring, how great is the 
tension in each cord? Ans. 4if lbs. ! 

n. Suppose that in the preceding example, the arc AB 
is a quadrant and AC and BC are equal. Determine the 
tensions. 

12. It is required to determine the weights necessary to 
hold a cord in the position shown in fig. 72(a). 

Solution: The lines radiating from (fig. 726) are drawn 
parallel to the segments of the cord, and AE is any vertical 
line. If the weights W lf W 2 , W 3 , and W 4 are proportional to 
AB, BC, CD, and DE respectively, they will hold the cord 
in the given position. For, imagine all the knots except the 




io6 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. 

first connected by vertical cords to a fixed support below, and 

a body hung from the first knot ; the 
cord is then in the required posi- 
tion. According to some scale, AB 
■-\ represents W v and according to the 
same scale BC, CD, and DE repre- 
sent the tensions in the imaginary 
vertical cords at the second, third, 
and fourth knots respectively. But 
these tensions can be supplied by 
bodies suspended from the knots. 

130. Position Assumed by a Cord 
Sustaining Loads. — Let fig. 73 repre- 
sent the nth. knot on the cord, the load suspended from which 
call W n and the tensions in the cord segments on the left 
and right T n and T n+1 respectively. Since 
the three forces represented are in equi- 
librium, — tF 
I 'F x =T n+1 co a n+1 -T n cos a n = o; T n 
IF y =T n+1 sin a n+1 — T n sm a n — W n = o. 
From the first equation, it follows that the 

horizontal components of the tensions in the cord segments are 
equal; that is, if H denotes the horizontal component in any 
segment, 

H=T n cos a n = T n+1 cos a w+1 = etc. . . . (1) 

This equation combined with the second one above gives 

tan a n+1 =tan a n + W n /H (2) 

By means of this equation the direction of each segment may 
be computed if, in addition to all the weights, the tension in 
and inclination of one segment or the inclinations of two ad- 
jacent segments are known. 




EXAMPLES. 



1. Three bodies weighing 100, 120, and 200 lbs. are sus- 
pended in the order named from three knots in a cord which is 
so supported that the second segment is horizontal and its 



II.] 



FLEXIBLE CORDS. 



107 



tension is 160 lbs. Determine the directions of the other seg- 
ments. Ans. a 1 = 32°. 

2. Suppose that the supports in the preceding example are 
such that the middle knot is the lowest and that the segments 
adjacent to that knot are inclined 30 above the horizontal. 
Determine the inclinations of the other segments and all the 
tensions. 

3. Solve the preceding examples graphically. 

131. The Loads are Equal and Uniformly Spaced Horizontally. 
— The knots are on a parabola; proof follows. Let O, 1, 2 . . . n 
(fig. 74) be knots on the cord from 
which the loads, W, are suspended, 
and let x and y axes be taken as in 
the figure. Denote the angles which 
0i, 12, 23, etc., make with the axes 
by a lt a 2 , a 3 . . . etc., the horizontal 
distance between consecutive loads 
by s, and the coordinates of the nth. 
knot by x and y. Then 

x = ns; (1) 

y = s (tan a^+tan a 2 + . . .). 

From eq. (2), art. 130, 

tan a 2 = tan a l + W/H ; 

tan a 3 = tan a 2 + W/H = tan a t + 2W/H 




Similarly, 



hence 



tan a 4 = tan a 1 + 7 ) W/H, etc. 



y = s[nta.na 1 + ^n(n-i)W/H] (2) 

Combining (1) and (2) and eliminating n, we get 



x 2 + 



2 tan a, H — W 



W 



2 Hs 



(3) 



which is the equation of a parabola. 

// the loads are so closely spaced as to constitute a practically 
continuous load, the cord is practically parabolic. The equa- 
tion of the parabola referred to vertical and horizontal axes 



108 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VL 

though the lowest point on it may be derived from eq. (3), or 
independently as follows: 

Let O (fig. 75) be the lowest point on the cord, Q any other 
y y fl point, and let w denote the load per 

unit horizontal distance; then OQ 
y sustains a load wx. The tension at 

_ x call H, and that at Q, T. The 
three forces applied to OQ are con- 
current, and their equilibrium equa- 
tions are 




ZF X =-H + T cos # = 0; . 
IF y = — wx + T sin # = 0. . 

Combining these with tan d = y + x/2, we get 

n 2H 



zv 




(1) 



(3) 



which is the equation sought. 

If the points of suspension are at the same level, then (see 
fig. 76 and eq. (3)), 

d = wa 2 /2H. ....... (4) 



EXAMPLES. 

i. A cord is supported at two points on the same level 30 
ft. apart, and its lowest point is 8 ft. below the level of the 
supports. If the load is 20 lbs. per ft., what are the tensions 
at the supports and at the lowest point ? Ans. H = 2&i\ lbs. 

2. Suppose that one support is 3 ft. higher than the other 
and that the lowest point of the cord is 8 ft. lower than the 
lower support. What is the greatest tension in the cord? 

132. Position Assumed by a Heavy Flexible Cord Suspended 
from Two Points. — The curve assumed by such a cord if uni- 
form in weight (and such only are discussed below) is called a 



§ II.] 



FLEXIBLE CORDS. 



109 



common catenary. Let AB (fig. 77) represent such a cord, C 
its lowest point, Q any other point, 5 the length CQ , H and T 




Fig. 77. 
the tensions at C and Q respectively, and w the weight of the 
cord per unit length. Then the forces on the portion CQ are 
H, T, and ws acting as shown in fig. 77(a); hence 

2F X = —H + T cos = o; 

IF y = — ms + T sin <£ = o. 
From these, by division, we get 

tan <f> = ws/H. 
For convenience, let c denote a length of the cord whose weight 
equals the tension at C, then H/w = c; and since tan <j> = dy/dx y 

%-h w 

Integration of equation (1) leads to the equation of the 
catenary. Differentiating, we get 

<£)-*• 

Since (ds) 2 = (dy) 2 + (<fc) 2 , ds = cfo Vi + {dy/dx) 2 ; 

hence 



or <i 

Integrating, we get 






k!+ 



Now dy/dx = 0, where # = (see fig.), hence C = o. 



no APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



Solving the equation for dy/dx, we find that 



(») 



e being the base of the Naperian system of logarithms. 
Integrating the last equation, we get 

y = -(e x/c + e-*/ c )+C". 

2 

Since y = c when x = o (see fig.), the constant C n is zero; hence 
the equation of the catenary referred to the axes of the figure is 



y = -(e x / c + e~ x / c ). 



(3) 



Several interesting relations may be deduced as follows 
From (i) and (2) 

e~*/ c ) 



S = -( e x/c 
2 



Combining (3) and (4), we find that 

y 2 = s 2 + c 2 . 



Also, 



y-\-s = ce*/ c , or x = c\og, 



y+s 



From the equilibrium equations, 

T 2 = w 2 (s 2 +c 2 ); 
hence T=wy. . . . 



(4) 

(s) 

(6) 
(7) 



EXAMPLES. 

i. A measuring-tape 400 ft. long weighing 0.005 lbs. per ft. 
is suspended from two points, A and 
B, the supporting pulls there being re- 
spectively 1.6 and 2.0 lbs. Compute 
the horizontal and vertical distances 
between A and B. 

Solution: Let T x and T 2 denote the 
pulls at A and B, and l x and l 2 the 
x lengths AC and BC respectively (fig. 78). 
Fig. 78. From eq. (7), 

T 1 = wb li or ^ = 1.6 -7-0.005 = 320 ft., 




§ II. J • FLEXIBLE CORDS. Ill 

and T 2 = wb 2 , or 62 = 2.0-7-0.005 = 400 ft.; 

hence b 2 — b 1 = 80 it. 

From eq. (6), a x = c\og e — — L , 

and a 2 = c log e — — - . 



To determine a x and a 2 values of l lt l 2 , and c are needed. 
Fromeq. (5), * 6 1 2 = / 1 2 -fc 2 = 102 400 

and 6 2 2 = / 2 2 -K 2 = i6o 000; 

also, / 1 -f/ 2 = 4oo. 

Solution of these three equations shows that 

^=128, ^ = 272, and (7 = 293-}- ft., 
which values if substituted in the expressions for a x and a 2 give 

a x = 122.7 and 03 = 239.9 ft.; 
hence a 1 + a 2 = 362.6 ft. 

2. What is the tension at the lowest point of the tape in 
ex. 1 ? 

3. A tape whose length is 100 ft. weighs 0.005 lbs. P er ft- 
and is subjected to end pulls of 10 lbs., the ends being at the 
same level. Compute the distance between the ends and the 
"sag." Ans. Distance = 99. 446 ft. 

Sag =0.38 ft. 

4. A tape is supported at two points on the same level, its 
length = 2/, its weight per unit length = w and the end pulls =P. 
If 2a and d denote distance between supports and the sag re- 
spectively, show that 

a=(P*/w*-l?)*log e P/w+l ± , 
d = P/w-(P 2 /u?-l 2 )*. 



1 1 2 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

133. Approximate Equations . — If the points of suspension 
are at the same level and the sag is small, the length of the cord 
and distance between supports are nearly equal, and the num- 
bered equations below are close approximations. They may 
be deduced as follows: 

Let 2/ denote the length of the cord and H the tension at 
the lowest point. From fig. 79 and eq. (6), art. 132, 

d=b — c and a = c\og e . 







Fig. 79. 
Fromeq. (s), c 2 = b 2 -l\ or c/b = (i-P/b 2 )±. 

From eq. (7), c = H/w and b = P/w; also, H = P cos a; 

hence c/b = cos a = ( 1 — l 2 /b 2 )^. 

Since a is very small l/b is also small; hence, expanding the 
binomial and neglecting the higher powers of l/b t 

c = b — l 2 /2b, approximately. - 

Substituting the value of b and the approximate one for c in 
the expressions for d and a above, gives 

d=wi 2 / 2 p. : (1) 

a-{P/w-wP/*P) log e p/ l^ /2p . . . (2) 

Or, we may reason thus: Since the slope of the cord is 
everywhere small, the load per horizontal unit length is nearly 
uniform; hence the catenary is very nearly parabolic and the 
equations deduced in art. 131 must be nearly correct for a flat 
catenary. 



$111.] TACKLE. 113 

From eq. (4) of that article, 

d = wa 2 /2H (3) 

Since a and I and H and P are nearly equal, the expressions for 
d in (1) and (3) are nearly equal. In works on calculus it is 
shown that the length of a parabolic arc, as OB (fig. 76), is 
given by 

/ = — tan a\/i + tan 2 a + log e (tan a + \/i +tan 2 a) , 

or, developed, 

. Hi tan 3 a \ 

/ = — tan a H K . . ). 

w\ 6 / 

Since in a parabola, tan a =wa/H, 

l = a(i+w 2 a 2 /6P 2 ), 

and a = l(i-w 2 l 2 /6P 2 ). ....... (4) 

EXAMPLES. 

i. Solve ex. 3, art. 132, by the approximate equations of 
this article. Ans. (1) gives for sag 0.625 ft. 

(4) " " distance 99.99 " 

2. Hard copper wire weighs 3.85A lbs. per foot, A being 
the area of its cross-section in square inches. If such a wire 
be suspended between two points at the same level 200 ft. apart, 
and the maximum pull on it be limited to 20000 lbs. per sq. in., 
what is the proper length of the wire and the sag? 

Ans. Sag = 0.96 ft. 

3. The rope of a rope drive when not running is observed 
to sag 1 ft., and the distance between the centres of the wheels 
is 70 ft. What is the maximum tension in the rope? 

§ III. Tackle. 

134. The Pulley. — A wheel with a flat-faced or grooved rim 
which can rotate about its axis is often used as a" power 
transmission ' ' device ; when so used it is called a pulley. It 
may rotate freely about an axle or shaft, and is then called a 



H4 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

loose pulley; or it may be fastened to a shaft which turns in 
bearings with the pulley, when it is called a fast pulley. 

A combination of one or more grooved pulleys and a rope 
or chain used for raising heavy bodies, is called a tackle. The 
pulley or pulleys rotating on the same axle together with the 
frame supporting the axle is called a block. A pulley is called 
movable or fixed according as the block of which it is a part 
does or does not move while the load is raised or lowered. 

135. Tension in a Cord on a Pulley. — Fig. 80 represents a 

^* *^v single pulley, loose or fast, and fixed or mov- 

t'i ^~x \ able, about a part of whose rim a flexible cord 

I v-^ J \ or belt is tightly wrapped. We wish to find 

x^ ^/ T " the relation between the tensions in the cord or 
Fig 80. bert on opposite sides of the pulley on the sup- 
position that the rubbing surfaces at the axle are ' ' smooth. ' ' 

If the rubbing surfaces at the axle are smooth, then the 
pressure, P, of bearings on the axle (fast pulley) or the pressure 
of the axle on the pulley (loose) is such that its line of action 
passes through the axis of the pulley (see art. 138). Then of 
the three forces T' t T", and P, the only ones having a moment 
about the axis are T and T" ', and if the cord or belt is flexible 
their arms are equal; hence T' = T". That is, the tensions in 
a cord or belt on opposite sides of a pulley which the cord or belt 
encircles are equal if the cord or belt is flexible and the rubbing 
surfaces at the axle are smooth. 

In the following examples this relation is made use of, the 
cords being assumed flexible and the rubbing surfaces smooth. 
This assumption is far from the truth in actual tackle, and it 
should be remembered that the results obtained below are for 
ideal cases. Axle friction and rigidity of cordage are discussed 
later. 

EXAMPLES. 

i. In the arrangements shown in fig. 81 (a) and (6), regard 
F and a as given and compute W and 0. 

2. Compute the pressure upon the axle (fig. 82), a and W 
being given. 

3. What is the relation between F and W in the tackle 



§ III.] 



TACKLE. 



"5 



shown in fig. 83, neglecting the weight of the parts? (Consider 
the equilibrium of the part below the dotted WMM r/ M 

line.) [ 

4. What is the relation between F and W 

the tackles shown in figs. 84 and 85, the 




Fig. 87. Fig. 86. 



weight of the parts being neglected ? What is the tension in 
the cord supporting the tackle of fig. 85? 

Ans. For fig. 85, W = 6F. 

5 . What is the relation between F and W in the tackle 
shown in fig. 86? What are the pulls on the hooks at A and B? 

6. Fig. 87 represents, in principle, a Weston differential 
tackle. The wheels in the upper block are fastened together, 
are of unequal diameters, and their grooves are made so that 
the hoisting chain will not slip in them. Determine the rela- 
tion between F and W, and the pull at the upper hook. 

Solution : Consider the equilibrium of the part of the tackle 
above the dotted line. The external forces upon it are four in 



n6 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

number if its weight be neglected, namely, the upward pull at 
the hook, the pull F, the tension in the chain on the left of the 
smaller wheel (W/2), and the tension in the chain to the right 
of the larger wheel (W/2). One condition of equilibrium for 
these forces is that their moment sum with respect to the axis 
of the wheels is zero ; hence if r' and r" denote radii of smaller 
and larger pulleys, 

Fr" + Wr'/2-Wr"/2=o, 

2Y" 

or W=-^— 7 F. 

7. Suppose that in fig. 81(6), a is zero and that a man whose 
weight is W sits upon the load and exerts a pull F. How 
great must the pull be to raise man and load? Is it greater or 
less than W? 

§ IV. Smooth Supports. 

136. Definitions. — It is known from experience that to slide 
one body over another even at constant speed requires the appli- 
cation of more or less force; also that a moderate "sliding 
force" may not cause a body to move. It is inferred that the 
second body exerts a force upon the first which is opposed to 
the sliding. 

More definitely, let fig. 88 represent two bodies whose sur- 

^ ^ p face of contact is a horizontal plane, the upper 

^/ j JV ? one being subjected to a horizontal force P. The 
i \r lower body exerts upon the upper a force such 

Fig. 88. bs R, the horizontal component of which, R x , is 
the resistance to sliding. If the upper body is in 
equilibrium, R X = P. Now it is known that the smoother the 
surfaces of contact, the smaller is the force required to cause 
sliding and hence the smaller the resistance to sliding. We are 
thus led to the conception of a 

Perfectly smooth surface, which may be defined as one which 
offers no resistance to the sliding of a body upon it. Such a 
surface is ideal, but there are surfaces which are nearly perfectly 
smooth. 



§ IV.] SMOOTH SUPPORTS. 117 

For simplicity, we will assume the surfaces of contact in 
some of the following examples as perfectly smooth. For 
brevity, they will be called smooth surfaces, while those whose 
resistance to sliding is to be taken into account will be called 
rough.* 

137. Reaction of a Smooth Surface. — A perfectly smooth 
plane surface can exert a force only along a normal to it. For, 
by definition, such a surface offers no resistance to the sliding 
of a body over it, that is, the reaction which the surface offers 
has no component along the surface and hence that reaction 
must be directed along the normal. If the surface is not a plane 
one, then the resistance exerted by each elementary part of 
the surface is directed along the normal to that part. 

EXAMPLES. 

1. A body weighing 100 lbs. rests upon a smooth plane 
inclined at an angle of 30 with the horizon, and is prevented 
from slipping by a cord fastened to it which leads off up the 
incline and over a smooth pulley at the top and supports a 
body which hangs freely from that end. Determine the weight 
of the suspended body and the resistance of the plane. 

Solutions: Consider the forces applied to the body upon 
the inclined plane ; they are three in number, namely, its weight, 
the pull of the cord, and the resistance of the incline. The 
system is coplanar and concurrent; and the pulley being smooth, 
the pull of the string equals the weight of the suspended body. 

(1) Algebraical. Let W denote the weight of the sus- 
pended body and R the resistance of the plane. 
The forces acting upon the body are represented 
in fig. 89. For such a system there are two 
equilibrium equations (art. 125), and if the x 
and y axes be taken horizontal and vertical re- 
spectively, 

IF X = W cos 3o°-R cos 6o° = o; 
IF y = W sin 30°+ R sin 6o° — 100 = 0. 




* Rough surfaces are discussed in § VIII. 




us APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VL 

These equations determine W and R. (The student should 
select the coordinate axes parallel and normal to the incline, 
write the equilibrium equations and compare the length of 
their solution with that of those above.) 

(2) Graphical. The condition of equilibrium is that the 

polygon for the three forces must 
close (art. 125). To construct the 
polygon , the wholly known force 
(the weight of the body) is repre- 
sented first; thus, AB (fig. 90) is 
drawn vertically and one inch long. 
Next, a line from A (or B) parallel 

lin.= 1001bs. tQ the incline> md Qne from B ( or 

A) parallel to the normal to the 
plane are drawn, thus determining C. Then BC and CA rep- 
resent the magnitudes and directions of the forces sought. 

2. Solve ex. (1), supposing that the pulley is so placed that 
the cord leads off horizontally from the body upon the inclined 
plane. Ans. W = 57 . 7 lbs. 

3. Solve ex. (1), supposing that the pulley is so placed that 
the cord leads off from the body upon the inclined plane at an 
angle of 50 with a horizontal. 

4. Two planes which are inclined at angles a and /? with a 
horizontal plane intersect in a horizontal line, and a cylinder 
whose weight is W rests between and upon them. Compute 
the pressure on each plane. 

5. A bar 24 in. long rests in a smooth hemispherical bowl 
30 in. in diameter. The centre of gravity of the bar is 10 in. 
from the lower end. Determine the position of equilibrium. 

Ans. Inclination of bar to horizontal is 12 32'. 
[Suggestion: The three forces maintaining the equilibrium of 
the bar are coplanar and concurrent. The action lines of two, 
the pressures of the bowl on the bar, intersect at the centre of 
the bowl and therefore the centre of gravity of the bar must be 
vertically below the centre of the bowl. Make a sketch show- 
ing this relation, mark the known lengths, and solve trigonomet- 
rically for the inclination of the bar.] 

6. A uniform bar whose length is 40 in. is supported by a 



§ IV.] SMOOTH SUPPORTS. "9 

smooth vertical wall and a smooth peg whose axis is horizontal, 
parallel to the wall, and 15 in. therefrom. Determine the posi- 
tion of equilibrium. Ans. Angle with vertical is 65 18'. 

138. Pin Joint or Hinge. — In some of the following exam- 
ples reference is made to bodies joined by means of a pin hinge. 
Such a joint may consist of a cylindrical pin and two cylindri- 
cal holes, one in each of the bodies joined, into which the pin is 
inserted allowing relative rotation about the pin. Still more 
simply, the joint may consist of a pin which is rigidly fastened 
to one body and inserted into a hole in the other, allowing 
rotation. 

If the cylindrical surfaces of the pin and hole are smooth, 
the forces exerted upon these surfaces act normally, and their 
resultant, therefore, passes through the axis of the pin. 

EXAMPLES. 

1. The body ABC (fig. 91) is supported by a smooth hinge 

at A and a smooth surface at B, and F 

a horizontal force F is applied at C. ~~% 



Compute the reactions of the supports, (-a --' — bO 

neglecting the weight of the body. "* 1 *i 

Ans, Reaction at B is Fa/ 1. 
2. Suppose that the support of the bar at B in ex. 1 is also 
a smooth pin hinge in the same horizontal plane with the other 
one. Determine the supporting forces. 

Solution : The reactions at M and N and the force F to be 
in equilibrium can not be parallel, hence they must be con- 

£_ current (art. 125); also their 



^c^,Jl--^'Zo T ^ **/£, force polygon must close. Any 

j jp— — %. reactions fulfilling these two con- 

ditions will balance F. Many 
/^Cj^^X 70 pairs of reactions can satisfy the 

A f b conditions — for example, BC and 

FlG - 92- CA, or BC and C'A (fig. 92); 

hence the problem is not statically determinate. 

The vertical components of the reactions are determinate, 
which may readily be proved thus: Call the reactions at M 



120 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

and N, R' and R" respectively, and imagine them resolved 
horizontally and vertically. The system F, R x , R y ', R x " , R y " 
is coplanar non-concurrent non-parallel, and there are three 
algebraic conditions of equilibrium (art. 125). They are 

IF x =F+R x '+R x " = o; 

ZF y =R y '+R y " = o; 
2Mtt=R y "l-Fa = o. 
From the last two equations it follows that 

R y "=Fa/l and RJ = -Fa/l. 
The indeterminateness, therefore, is really with the horizontal 
components; their arithmetic sum or difference equals F, as 
may be seen from the first equation or from the force polygon 
in fig. 92. 

3. A uniform bar weighing 200 lbs. and 10 ft. long leans 
against a smooth vertical wall, and its lower end is fastened 
to a floor by a smooth pin hinge whose axis is horizontal and 
parallel to the wall. The distance of the hinge from the wall 
being 8 ft., determine the forces supporting the bar, solving 
graphically. 

Ans. Force on upper end is horizontal and equals 133 J lbs. 

§ V. Three Typical Problems on Coplanar Non-concur- 
rent Forces. 

139. Problem I. — The forces of a parallel system in equi- 
librium are completely known except two whose action lines only 
are known. It is required to determine completely these two forces. 

Algebraic Solution. — There are two conditions of equilibrium 
for this system (art. 125), namely, 

IF =0 and IM =0, 
or IM a = o and IM b =0.* 

Either set of equations furnishes the solution. It is advan- 
tageous to select moment origins on the lines of action of the 

* Whenever a force whose sense is unknown is to be entered into a 
resolution or moment equation, a sense should be assumed for that force 
and adhered to in the solution of the equation. The correct sense is 
indicated by the sign of the computed value of that force. It is as 
assumed or opposite according as the sign is positive or negative. 



§v.] 



THREE TYPICAL PROBLEMS. 



121 




Fig. 93. 



unknown forces. Then each moment equation will contain 
but one unknown quantity and may be readily solved. 

Illustration: Let it be required to determine the magni- 
tudes and directions of the two forces 
F' andi 7 " (fig. 93), all the forces there 
represented being in equilibrium, F t , 
F 2 , and F 3 being 700, 300, and 500 lbs., 
and a, b, c, and d 1, 3, 5, and 2 ft. 
respectively. 

If F r and F" are assumed to act upward and the moment 
origin is taken on F" ', 

IF= — joo+F' — 3oo + 5oo + i 7 " = o; 
IM= +700-11— F f -10 + 300 -7 — 500- 2=0. 

From the second equation, ^ = 880 lbs., and this value sub- 
stituted in the first gives F" = — 380 lbs. The minus sign 
means that F" does not act up as assumed, but down. 

Employing the second set of equilibrium equations, and 
selecting moment origins on F' and F" respectively, 

.JM= 700-1 — 300- 3 + 500-8 + F"- 10 = 0; 
2'M=7oo- 11 —F'- 10 + 300-7— 500-2 =0. 

These solved give the same values as those found above. 

Graphical Solution. — The conditions of equilibrium are that 
the force and funicular polygons close (art. 125). In the 
process of constructing them the unknown quantities will be 
determined. 

Illustration: Let the data be the same as that in the pre- 
ceding illustration. First the force polygon should be drawn 
as far as possible; thus, AB, BC, and CD (fig. 94) representing 
F lt F 2 , and F 3 respectively. If either one of the unknown 
forces be lettered DE, the other must be FA because the poly- 
gon must close. It remains to locate the point E; this is done by 
means of the funicular polygon. Before beginning to draw the 
polygon, recall what the strings represent (see arts. 37 and 38). 
If the polygon be begun on ab, strings oa and ob must be drawn 
from that point ; then oc is drawn from where ob intersects be, 
and od from where oc intersects cd. Now oa is the action 



122 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

line of one of the components of EA, and should be extended 
to ea t thus determining a point / in the action line of the other 
component, EO. Also, od is the action line of one of the com- 




Scale:lin.=6ft. 



Soale.lin.= 1001bs. 



(a) (b) 

Fig. 94. 

ponents of DE, and should be extended to de, thus deter- 
mining a point II in the action line of the other component, 
OE. Now, if 'the funicular polygon is to be closed, the action 
lines of EO and OE must coincide (see art. 40) ; hence there is 
but one string oe, and it passes through points / and II. The 
ray corresponding to oe may now be drawn, thus determining E. 



EXAMPLES. 

1. Reverse F x and F 3 (fig. 93) and solve algebraically the 
example used for an illustration above. 

2. Solve the preceding example graphically. 

140. Problem II. — The forces of a non-parallel system in 
equilibrium are completely known except two. Of these, the action 
line of one and a point in that of the other are known. It is re- 
quired to determine completely these two. 

Algebraic Solution. — There are three conditions of equi- 
librium for a system of this kind (art. 125), namely, 
IF x = o, IF y = o, IM =0; 
IF x = o, IM a = o, IM b = o; 
or, IM a = o, IM b = o, IM c = o. 

Either set furnishes the solution, the three unknowns being 
the magnitude (and sense), of one force, the magnitude (and 
sense) and inclination of the other. 



§V.] THREE TYPICAL PROBLEMS. 123 

Illustration : Suppose a horizontal beam which is supported 
at one end by a pin hinge and at the other by an inclined cord 
sustains given loads, and that it is required to determine the ten- 
sion in the cord and the pin reaction. 

Let fig. 05 represent the beam, loading, notation etc.; the 
p 

ft /\Hinge D C E__ 



TT 



_ 2 i — **--]«.- 



50 lbs. 100 lbs. 25 lbs. 
Fig. 95. 

100-lb. force may be the weight of the beam. The unknowns 
are denoted by Q, P, and 6* Employing the first set of equi- 
librium equations and selecting the x and y axes horizontal and 
vertical respectively, and a moment origin at .4, 

IF X = -Pcos6> + Qcos6o° = o; . . . . . . (1) 

IF y = P sin 6 + Q sin 6o° — 50 — 100 — 25 =0; . . (2) 
IMa = Q-6 sin 6o° — 50-2 — 100-3 ~ 2 5 ' 5 = °- • • (3) 
These equations furnish the solution. Observe that by select- 
ing A as moment origin, two unknowns, P and 6, were elimi- 
nated from the moment equation. 

Usually it is , advantageous to replace the force whose 
magnitude and direction are unknown by two rectangular 
components acting at the known point in the action line of 
the force. Then the unknowns, instead of P and 6 of the illus- 
tration, would be P x and P y , and when these latter become 
known, P and d are easily computed. Such substitution trans- 
forms the problem to Problem III (art. 141). The example 
employed above as illustration is solved in art. 141 on this plan. 
Graphical Solutions. — (1) The conditions of equilibrium are 
that the force and funicular polygons close (art. 125). In the 
process of constructing the two polygons the unknowns will 
be determined. 

* In this example it is evident that the senses of P and Q are as repre- 
sented and that #<9o°. When the senses of P and Q and "quadrant" of 
d are not evident, they should be sketched as they appear to be, and then 
the equilibrium equations should be written. The correct senses may 
be inferred as explained in the foot-note, p. 120, and the quadrant of 
will appear from the solution. 



124 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 




Scale:lin. = 2001bs, 

(b) 

Fig. 96. 



Illustration: Let the data be the same as in the preceding 

illustration. First, the force poly- 
d / gon is drawn as far as possible; 
d/ e thus, AB, BC, and CD (fig. 96), 
■1 fifc^yOn representing the 50, 100, and 
25 lb. forces, and then an in- 
definite line parallel to the cord 
from D (or A ) . The fourth force, 
tension, then is to be lettered 
DE, and the remaining one must 
be marked EA, since the poly- 
gon is to be closed. It remains 
to determine E; this is done by 
means of the funicular polygon. 

If the polygon be begun at 
any point of ab, be, ed, or de, 
it can not be completed, as is plain from an examination 
of the unlettered funicular polygon which was begun from a 
point of ab taken at random. The difficulty is in determining 
the desired intersection of the string corresponding to OA and 
the unknown action line of EA ; this point corresponds to 
point I of fig. 94. If the funicular polygon be begun at the 
given point of the unknown action line, the difficulty is avoided; 
thus, oa is drawn through that point, and then ob, oc, and od. 
Now 7 and II respectively are points in the action lines of the 
forces EO and OE, and if the funicular polygon is to be closed, 
the lines of action of EO and OE must coincide (art. 45); hence 
there is but one string oe, and it passes through points I and 
II. The ray corresponding to OE may now be drawn, and thus 
the point E is determined. 

(2) Find the resultant of the wholly known forces, and 
consider the system as consisting of that force and the two 
unknown ones. Then, making use of the fact that the three 
forces are parallel or concurrent (see art. 125), determine the 
unknowns.* 

* If the resultant of the known forces is a couple, the two unknowns 
must also constitute a couple. How might one determine graphically 
the forces of the latter couple ? 



I v.] 



THREE TYPICAL PROBLEMS. 



125 



Illustration: Let the data be the same as that of the pre- 
ceding illustration. The resultant of AB, BC, and CD (fig. 





a 


b b 


c c 


d 


IV GZ 






/■j 








// 


t 


\\ V 


t "' T J 








/ 




n n 


' / 


V ^/o 


\ / 


/ // 


\ ad / 


/ ' / 


\ 




1 / 


\ 
\ 


/ 

/ 




A 




f I / 


\ 


/ 


//' Scale: 


v ' Scale: 


' lin.=1201bs. 






liiu=4ft. 



(b) 



(a) 



Fig. 97. 



97) is found to be AD. Extending ad and de we find 
their intersection which is one point in the action line of 
the other unknown force; another point in its action line 
being given, the centre of the pin, the line is known. The 
action lines, ad, de, and ea, of three forces in equilibrium 
and the magnitude and sense of one, AD, are now known. To 
determine the remaining elements, we have only to draw the 
force triangle for the three forces, ADE. Then DE represents 
the magnitude and sense of the pull of the cord, and EA the 
magnitude and sense of the pin pressure. 



EXAMPLE. 

Suppose that the bar represented in fig. 98 is supported by 
a smooth surface at A (then the reaction there 
must be horizontal) and by a pin joint at B. 
Determine the supporting forces graphically. 

141. Problem III. — The forces of a non- 
parallel system in equilibrium are completely known 
except three whose lines of action only are known. 
It is required to determine completely these three. 

Algebraic Solution. — There are three conditions 
of equilibrium for a system of this kind, as in the preceding 




126 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI 

article. Either set of equations determines the unknowns, the 
magnitude (and sense) of three forces. 

Illustration: Let the data be the same as that of the illus- 
tration of the preceding article, but imagine P replaced by its 
components P x and P y as in fig. 99. 

p x *!* , j— r-^Oeoo 

50 lbs. 100 lbs. 25 lbs. 
Fig. 99. 

(a) Employing the first set of equilibrium equations, 

IF X = ~P X + Q cos 6o° = o; (1; 

2Fy = P y + Q sin 6o° — 50 — 100 — 25 = 0; ... (2) 
IMa = Q-6 sin 6o° — 50- 2 — 100-3 — 2 5 ' 5 = °- • • (3) 

Equation (3) is just like eq. (3) of the preceding article. The 
value of Q determined from (3), substituted in (1) and (2) 
leaves them with two unknowns, P x and P y . Now it is con- 
siderably easier to solve (1) and (2) for P x and P y than to solve 
(1) and (2) of the last article for P and d, as may be seen by 
trial. The reason lies in the fact that both sin 6 and cos 6 
appear in those equations, and one really has to solve three 
equations, the third one being sin 2 #+cos 2 = 1. 

(6) Employing the second set of equilibrium equations, 

IF X =-P X + Q cos 6o° = o; . (4) 

2Ma = Q'6 sin 6o° — 50- 2 — 100-3 ~~ 2 5 ' 5 — °t • • (5) 
IMb= -iV6 + 5o-4 + ioo-3 + 25-i=o. ... (6) 

Equations (5) and (6) contain but one unknown each; no elimi- 
nation therefore is necessary to obtain P y and Q. The value of 
Q substituted in (4) leaves but one unknown in that equation. 

(c) The third set of equilibrium equations may be written 
for the forces of this example so that but one unknown will 
appear in each equation, and then no elimination is necessary 
in solving the equations. The student should so write them. 

Graphical Solutions — (1) If one imagines any two of the 
unknown forces replaced by their resultant, the problem is 
transformed to Prob. II (art. 140). Thus, let F', F" , and F'" 



v.] 



THREE TYPICAL PROBLEMS. 



127 



denote the three unknowns, and R the resultant of any two of 
them, F" and F'" say; then in the transformed system there 
are two partially unknown forces F' and R. The action line. 
of F f is known, also one point in that of R, the intersection 
of the action lines of F" and F'" . Therefore, by the method 
of Prob. II, F' and R may be determined, and then R may be 
resolved into two components whose action lines are those of 
F" and F'" respectively; these components are F" and F"\ 

Illustration: It is required to determine the supporting 
forces, "reactions," on the overhanging framework of fig. 100 
when loaded as shown. The sup- 
ports are such that the reaction 
above is directed along the rafter, 
and reactions below are horizontal 
and vertical. Imagine the lower 
two reactions compounded and 
call the resultant R; its line of 
action is unknown, but it passes 
through the lower point of support. 
Then the system in equilibrium 
consists of R, the upper reaction, 
and the loads. The force polygon 
for the known forces is ABDCE, 
and since the direction of the 
upper reaction is known, the poly- 
gon can be continued by a line 
from E parallel to the upper reaction. The as yet unknown end 
of that line is to be marked F, and then the force R will be 
FA, since the force polygon is to be closed. 

The strings which intersect on af are oa and of (art. 38), and 
since one point on af, P, is known and also OA, the string oa 
may be drawn. Continuing the construction, ob is drawn next, 
then oc, od, and oe. Now / is a point in of and P is another 
hence of may be drawn and then the corresponding ray. That 
ray determines F, and then EF and FA represent the uppe - 
reaction and R respectively, and the horizontal and vertical 
components of FA (FG and GA) represent the supporting 
forces at P. 




Fig. 100. 



128 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



Scale: lin.- 2400 lbs. 



(2) Find the resultant of the wholly known forces; then 
the system in equilibrium may be regarded as consisting of 
that resultant and the three partially unknown forces. The 
special condition of equilibrium for four such forces is that 
the resultant of either pair balances that of the other pair 
(art. 121). 

Illustration: Let the data be the same as that of the pre- 
ceding illustration, and let F denote the upper reaction and 
H and V the lower ones (fig. 101). The resultant of the loads 

is a force of 1500 lbs.; its action line 
is ab. Let R' denote the resultant of 
the pair F, 1500, and R" that of 
the pair H, V. Now R f passes 
through Q and R" through P, and 
since they are collinear, PQ is the 
action line of each. Since the three 
forces F, 1500, and R" are in equilib- 
rium, and since their action lines and 
the magnitude and sense of one are 
known, their force polygon can be 
drawn; it is ABC A, BC representing the magnitude and direc- 
tion of F. Since the four forces, 900, F, H, and V, are in 
equilibrium, their polygon must close; hence draw lines from 
A and C parallel to H and V determining D. Then ABCDA 
is the force polygon sought, and CD and DA represent the 
magnitudes and directions of V and H respectively. 

EXAMPLES. 

1. Suppose the truss represented in fig. 102 to be supported 




Fig. ioi. 




on a smooth surface at A and to be pinned to the wall at B. 
Determine the reactions at A and B due to the loads, by both 
methods. 



VI.] 



JOINTED FRAMES. 



129 



2. Determine the reactions on the crane in fig. 103 due to 
the load and weights of members by both methods. (The 




brace 1.1 



Fig. 103. 

supporting surfaces are such that the upper reaction is hori- 
zontal and below one is horizontal and one vertical.) 



§ VI. Jointed Frames. 

142. Definitions. — A plane framework is one all members of 
which are parallel to a plane ; only such frames are considered 
in this section. A jointed frame is one the members of which 
are fastened by pin-joints, the pin being perpendicular to the 
plane of the frame. It is assumed in the following that the 
pins are smooth. 

143. The Pin Pressures. — The simplest kind of a member 
is one which is straight and is joined to others at its ends (fig. 
104). Such a member, if sustaining no load, is subjected to 




Fig. 104. 

three forces, its weight and the pin pressures, W, P', and P" 
respectively. Unless the member is vertical, the action lines of 



13° APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

the pin pressures do not coincide with the axis of the member; for, 
assuming that they do, it is plain that P' and P" cannot bal- 
ance W. If the weight of the member is neglected, the two 
forces P' and P" act along the axis and are equal; but, in 
general, a pin pressure is not directed along the axis of the 
member on which it acts. 

144. General Direction for Solving Examples. — The student 
should read again art. 127. In the following examples it will 
not always be evident to the beginner which body and which 
force system to consider for determining any particular force. 
Often several may be selected in the manner mentioned in the 
article referred to, and it may be that there is but little choice 
between them; but, as a general rule, the body should, if pos- 
sible, be so selected that the number of unknown elements in 
the external system of forces acting upon it shall no' L , exceed 
the number of conditions of equilibrium for that system. 

EXAMPLES. 

1. Determine the forces upon each member of the crane in 
fig. 103, neglecting their weights and taking x equal to 16 ft. 

Solutions: (1) Algebraic. Fig. 105 represents the whole 
crane, its members and groups of them, and the corresponding 
external forces. There are four external forces applied to the 
crane, namely, the load, and the reactions X, Y, and H (fig. 
105a). The brace (fig. 1056) is subjected to two forces only; 
hence they are collinear, and each acts along AC. The boom 
(fig. 105c) is subjected to three forces, — one at C, one at B, 
and the load ; the one at C is the reaction corresponding to the 
pressure on the upper end of the brace, and hence is collinear 
with that pressure. The force at B is unknown in direction, 
and is therefore represented by its components B x and B y . 
The post (fig. 1 05 d) is subjected to five forces, — X, Y, H t one 
at A, and one at B\ the latter two are reactions corresponding 
to the forces upon the left ends of the boom and brace respect- 
ively. 

An examination of these systems reveals several orders of 
procedure. In general, it is advisable to determine the reac- 
tions on the entire frame first, if possible; here it is possible, 



§ VI.] 



JOINTED FRAMES. 



13* 



for the system applied to the entire crane is coplanar non- 
concurrent non-parallel with three unknowns. Their deter- 
mination is left for the student; the values for X, Y, and H 
are 7.1 1, 8, and 7.1 1 tons respectively. System (c) might be 

-h 




Fig. 



105. 



solved next, for it is coplanar non-concurrent non-parallel 
with three unknowns. The student should prove that B Xf B y , 
and C equal respectively 10.67, 1.14, and 14.05 tons. From 
system (6) it is plain that A equals C. Finally, 5 = 10.73, 
and the inclination of B with the horizontal is tan -1 B y /B x = 
6° 7'. All the unknowns have now been determined. 

The student should examine the remaining force systems 
represented in fig. 105, and determine other orders of procedure 
for determining the unknowns. 

(2) Graphical. The system of external forces applied to 
the crane may be solved like Prob. Ill (art. 141). The force 
polygon for the four forces is ABCDA (fig. 106a), AB repre- 
senting the load. The forces applied to the boom, being three 
in number, are concurrent; hence their action lines are known. 
Since the magnitude of one force is known, their force polygon 
can be drawn, thus determining the magnitudes of the other 
two. The polygon is ABE A (fig. 106b \ AB representing the 
load. All the unknowns have now been determined. 



132 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



Instead of solving the force system applied to the boom, we 
might have solved that applied to the mast. There are five 
forces in that system, — three wholly known (BC, CD, and DA), 
the action line of one, and a point in that of the last. The 
resultant of the three wholly known forces is AB (fig. 106c); 





Scales, of space diagram 1 in.= 20 ft. 

"vector " lin.= 20tons. 

Fig. io6. 

and if the three known forces be imagined replaced by their 
resultant, then the system consists of three forces, and it 
must be concurrent. The action line of the last force is now 
known since it passes through the action lines of the result- 



Load 




Fig. 107. 



Fig. 108. 



ant and the fourth force, and the force polygon for the three 
forces may be drawn, thus determining the remaining un- 



§VL] 



JOINTED FRAMES. 



133 



knowns. The polygon is ABE A, BE and EA representing 
the fourth and fifth forces. 

2. Include the weights of the members in ex. 1 and solve. 
(Notice that the forces applied to the ends of the brace are not 
directed along its axis. It will be convenient in an algebraic 
solution to replace each unknown force by its horizontal and 
vertical components.) 

3. The weights of the post and jib of the crane represented 
in fig. 107 are 1 and f ton respectively; that of the tie may be 
neglected. Take the load as 8 tons and xas 10 feet, and com- 
pute all the forces on each member by both methods. 

4. The weights of the post and jib represented in fig. 108 
are 1200 and 1500 lbs. respectively; that of the tie may be 
neglected. For a load of 4 tons compute all the forces upon 
each piece by both methods. Disregard 
counterweight, shown dotted. 

5. Solve the preceding example on 
the supposition that the crane has a 
counterbalance whose weight is 10 tons, 
its centre of gravity being 9 ft. from 
the axis of the post. 

6. Determine all the forces upon 
each member of the crane * repre- 
sented in fig. 109 due to a load of 3 
tons 7 ft. from C. 

7. Fig. 110a represents a type of hydraulic crane. The 
plunger works inside a hollow mast and, pressing against the 
bottom of the boom, raises and lowers the load and boom to- 
gether. Compute the forces upon each piece due to a load of 
10 tons, x feet from the axis of the plunger when the pin A is 
y feet above the floor. 

Solution: Consider first the entire crane (fig. 11 06), except 
post and plunger. The external forces applied to it consist of 
the load, the post pressure upon the upper roller, R lt the post 

* It is impossible to analyse this crane by the preceding principles. 
To make it possible, suppose that the hole in the mast at C is slotted as 
shown (then the pressure there is vertical), and that there is no member 
AE. 




134 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

pressures on the lower rollers, R 2 each, and the plunger pres- 
sure P. The three unknowns may be determined, for the sys- 




FlG. no. 

tern is coplanar non-concurrent non-parallel. The solution 
gives 

^ = 1.433;, R 2 = o. , jix, and P=iotons. 

To continue, we may consider next the pin and roller at B. 
The external forces upon this body are five in number, the pres- 
sure on the roller 1 .433;, the forces exerted on the pin by the two 
members AB and the two BC. The four pin pressures are 
directed along the axes of corresponding members (fig. hoc). 
Solution of the equations of equilibr um of this system gives 

F' = o.2$x and .F" = 0.843; tons. 

Upon the pin at A there are applied five forces, two by the 
members AB, two by the rollers, and one by the boom; their 
magnitudes are respectively 0.263;, 0.713;, and 1.673;. The 
student should prove the values given. 

8. The frame of fig. 1 11(a) rests upon smooth surfaces at 
A and B t Determine the forces (pin pressures) upon each 
member. 



§ VI J 



JOINTED FRAMES. 



J 35 



Solution: From a consideration of the external forces on 
the collection of bars it follows readily that the forces at A and B 
are 446 and 554 lbs. respectively. Fig. in (6), (c), and (d) rep- 
resents the external system on each member, each pin pressure 
being replaced by its horizontal and vertical components. No 
unknown of the system (b) or (c) can be determined from 
the equilibrium equations of the system, but the forces D y 




D * to, * s 

y 1000 lbs. 

Fig. hi. 



and E y may be computed from the equations for system (d). 
Making use of these values in (a) and {b), the remaining 
unknowns can be determined. The student should make the 
determination. 

9. The bars of the frame of fig. 112(a) are uniform, AC, BC, 
and AB weighing 150, 100, and 200 lbs. respectively. Com- 
pute the forces upon each bar. 

Solution: The forces applied to each member are repre- 
sented in fig. 112 (6), (c), and (d), each force whose direction is 
unknown being represented by its horizontal and vertical com- 
ponents. The senses of the vertical components at C are not 
obvious, so they are assumed (see foot-note, p. 120). 

From a consideration of the external system on the entire 



I3 6 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

frame, R x and R 2 are readily found to be 229 and 221 lbs. re- 
spectively. From the system (d), A y and B y are found to be 
129 and 121 lbs. respectively. Supplying the value of A y in 
system (6) or B y in system (<:), the remaining unknowns may 
be computed. 




Instead of following the order above, one might write the 
equilibrium equations for systems (b) and (c) and solve them 
for the six unknowns, and lastly determine R 1 and R 2 . 



§ VII. Jointed Frames (Continued). 

145. Kind of Frames Considered. — The jointed frames con- 
sidered in this section differ from those of the preceding section 
in construction and in loading. It is assumed that 

(a) each member connects only two joints, 

(6) each load is applied so that its action line passes through the 
axis of a joint. 

146. "Force or Stress in a Member." — Fig. 113(a) repre- 
sents a member of such a frame as described in the preceding 
article, under two loads U and L". The pin pressures are 
denoted by P f and P", and the weight of the member is 
neglected or not considered. Let the resultants of the forces 



§VIL] JOINTED FRAMES. 137 

at the left and right ends be denoted by R' and R" respect- 
ively; the action line of each passes through the centre of 
the corresponding hole. Since R' and R" balance, they must 
be collinear, and their action lines must coincide with the axis of 
the member. 

Now any two parts of the member, as M and N, exert forces 
on each other, and the lines of action of those forces coincide 



3 



Q M_ 



**^(a> 



<- + M ■ h » <- l N ^ (b) 



-B ^ m H <- - H n 4 ^' (o) 

Fig. 113. 

with the axis of the member. For, the fdrce which M exerts on 
N balances R" , therefore its action line must coincide with that 
of R"; also the force which N exerts on M balances R' , there- 
fore its action line must coincide with that of R r . 

Observe carefully the relations in fig. 113 (b) and (c). 
In (b), the forces at the section between M and N are pulls, 

or the stress is tensile (art. 104), and the member is stretched 

by Rf and R". 
In (c), the forces at the section between M and N are 

pushes, or the stress is pressural (art. 104), and the member 

is compressed by R' and R". 

By force or stress in a member is meant either of the forces 
which a part of it exerts upon the other. The examples in this 
section relate to the determination of the forces in the members 
of jointed frames. 

147. Method for Determining the Force or Stress in a Member. 

(1) Determine the reactions on the frame, or truss. 

(2) Imagine the truss separated into two distinct parts* so 

* Such division of a truss is also described as 'passing a section," the 
term section referring to the imaginary surface of separation. 



I3 8 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

that the member under consideration is one of the members 
separated, and so that the unknown elements in the system of 
external forces applied to one part of the truss does not exceed 
the number of algebraic conditions of equilibrium for that 
system. 

(3) Apply the appropriate condition or conditions of equi- 
librium necessary to determine the desired force. 

The student should bear in mind that the system of external 
forces with reference to any part of a truss consists of the loads 
and reactions applied to that part and the forces which the 




SI 







other part exerts upon it. These latter forces are exerted upon 
the ' ' cut ' ' ends of the members belonging to the part considered 
and are exerted along their axes. 

Illustration 1. — It is required to determine the force in 
the member AB of the truss of fig. 114(a). 

The reactions are supposed to have been determined. If 
the section cutting AB be passed as at 2, the external system 
on either the upper or the lower part of truss (fig. 1140) includes 
four -unknown forces, those in the four members cut. Now 
the former system is concurrent and since it has only two con- 
ditions of equilibrium the unknowns cannot be determined 



§VIL] JOINTED FRAMES. 139 

from that system. The system on the lower part is non-con- 
current, and as it has only three conditions the unknowns can- 
not be determined from it. 

If the section be passed as at II, the system on each part of 
the truss contains only three unknown forces (fig. 114c); and 
since each system is non- concurrent, there are three conditions 
of equilibrium, and the force desired can be obtained from either 
system. 

Illustration 2. — It is required to determine the stress in 
member BC , fig. 114(a). 

No matter where the section cutting CB is made, the ex- 
ternal system on either part will contain more unknown forces 
than the number of conditions of equilibrium for the system, 
and the desired stress cannot be so directly determined. Thus, 
if the- section is made as at II L, the system on the left part is as 
represented in fig. 114(d), and there are four unknowns, the 
forces in the four cut members. If now the force in CD, BE, 
or EF can be determined, its value may be supplied in fig. 114(d) 
and the system can then be solved for the desired force, for there 
are three conditions of equilibrium and but three unknowns. 
The force in CD can be determined from the system of fig. 
114(c). 

When the stress in each member of a truss is required, a 
certain order of determining them, depending on the case in 
hand, is more convenient than others; but in all cases, the sec- 
tions are made according to direction (2) as stated above. It 
does not fall within the scope of this book to explain fully the 
most convenient orders of procedure for the different cases. 
A good general method is to make the sections so that the ex- 
ternal system on one of the parts of the truss shall be simple, 
containing few unknowns and easy to solve. This matter is 
partially illustrated in the solution of the first of the following 
examples and in arts. 148-151. 

EXAMPLES. 

1 . Determine the force in each member of the frame of 
fig 115(a) due to the load of 1000 lbs 



140 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

Solution : Imagine the truss divided into two distinct parts 
as shown in fig. 115 (6) and (c). The external forces upon 
the left part are 400 lbs., F x ' and F 2 f ; upon the right part 600 
lbs., 1000 lbs., F x " andF 2 ".* Each is a system in equilibrium, 
and the unknowns may be determined from the equilibrium 
equations for either system. 

wool lbs. 




*y 



^* (*) 



^400 lbs 



600 lbs 



A 



Fig. 



To determine the force in the member BC, the truss should 
be imagined as separated into two parts so that BC is one of 
the members cut, for example as in fig. 115 (d) and (<?)• The ex- 
ternal forces on the left part are 400 lbs., 1000 lbs., F 2 ' and F 3 '; 
upon the right part 600 lbs., F" and F 3 ". Each is a system in 
equilibrium, and the equilibrium equations for either determine 
the remaining unknown F 3 . 

*F l f ^F l " 9 F 2 f ^F i " 9 etc. 



§VIL] JOINTED FRAMES. 141 

In fig. 115 (/) and (g), there are represented the external 
systems on the parts of the truss made by cutting members AC 
and BC. The equilibrium equations for either determine i 7 , 
and F 3 . 

Of course, consideration of all the systems represented is 
not necessary for a solution. They are referred to here merely 
to show that the solution may be made in several different 
ways. One of these ways is by means of the systems of fig. 
115 (e) and (/), which may be carried out thus: 

Equilibrium equations for system (e) are 

IF X = F 3 " cos 3 6° 52'-F 2 " = o; 
IF y = -F 3 " sin 36 52 r + 6oo = o;* 



hence 



FJ r = 1000 and F 9 " = 800 lbs. 



Having determined F 3 ", F 3 ' is known; it is a push, that is, it 
acts upward (fig. 115/) and its value is 1000 lbs. Only one 
unknown remains in system (/), and the following equilibrium 
equation suffices for its determination: 

IF x = F l " cos 26 34' — 1000 cos 36 52' = 0; 
hence ^" = 894.5 lbs. 

Supposing the senses of the unknown forces in the two 
systems just considered not apparent, and following the sug- 
gestion of the foot-note, the first two equations above become 

-F 3 "cos 3 6 52'-F 2 " = o; 
F 3 "sin 36 52 r + 6oo = o; 

from which F 3 " = — 1000 and F 2 " = + 800 lbs. 

* In simple trusses the kind of stress in any member is apparent. 
For example, in Fig. 115(a), AC and BC are in compression and AB in 
tension; then F t and F 3 are pushes and F 2 is a pull. When the senses of 
the forces are not apparent, we may follow the suggestion in the foot-note, 
p. 120, but it is convenient to always assume the force to be a pull. Then, 
according to the foot-note, the force is actually a pull or push {and the member 
is in tension or compression) according as its computed value is positive or 
negative. 



142 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

Interpreting these signs in accordance with the foot-note, F" 
is a push and F 2 " a pull, i.e., BC is in compression and AB in 
tension — results agreeing with the first solution. 

2. Determine the force in each member of the truss of 
fig. 116. Ans. A F, 1600 lbs. tension. 




Fig. 116. 





3. Determine the force in each 
member of the truss of fig. 117. 

Ans. AD, 2600 lbs. compression; 
AC, 1300 lbs. tension. 

4. Determine the force in each 
b member of the truss of fig. 118. 

5. Determine the force in each 
member of the truss of fig. 102, it being 

supported as there described. 

148. Graphical Method for "Analyzing Trusses." — Graph- 
ical methods are especially well adapted for solving problems 
like the preceding. As in the algebraic method, the truss is 
imagined separated into two parts and then the attention is 
directed to the external forces acting upon either part. Graphi- 
cal instead of algebraical conditions of equilibrium are then 
applied to the system of forces to determine the unknowns. 
In making the imaginary separations of the truss, care should 
be taken to cut not more than three members, the forces 
in which are unknown. It is advantageous to make the separa- 
tion so that not more than two such members are cut. If that 
be done, a single force polygon will determine the two unknowns, 
while if three be cut, a force polygon and a funicular polygon, 
or the equivalent, are necessary for determining the three un- 
knowns. 



§ VII.l JOINTED FRAMES. 143 

149. Notation. — The notation described in art. 11 when 
applied in the graphical analysis of trusses can be advanta- 
geously systemized as follows. Each triangular space in the 
truss diagram is marked by a small letter, also the space 
between consecutive action lines of the loads and reactions 
(see fig. 119) Then the two letters on opposite sides of any 
line serve to designate that line, and the same large letters are 
used to designate the magnitude of the corresponding force. 
This scheme of notation is a great help in graphical analyses of 
trusses. 

Illustration. — Determine the force in each member of the 
truss of fig. 119. 

Solution: Evidently the reactions each equal one-half the 
load, or 2000 lbs. Imagine the truss separated into two parts, as 




by the arc I. The external forces upon the left part are repre- 
sented as far as known in fig. 120(a); since they are in equilib- 
rium, their polygon closes, and in constructing it, the unknowns 
will be determined. Beginning with the knowns, AB is drawn 
to represent 2000 lbs., BC to represent 500 lbs.; and then a 
line from A (or C) parallel to the action line of one of the un- 
knowns and a line from C (or A) parallel to the other are drawn. 
The last two lines determine D (or D'), and the closed polygon 
is ABCDA (or ABCD'A); hence the forces in the members cd 
and ad are represented by CD and DA (3000 and 2600 lbs.) 
respectively. From the force polygon, it is seen that CD is a 
push, and DA is a pull; hence the members cd and ad are in 
compression and tension respectively. 

We may next imagine the truss separated into two parts 
as by II or IV (fig. 119); in either case, there are but two un- 



144 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

known forces in the external system applied to either part. If 
we choose the part within II, we have the simpler system to 
deal with; the forces of it are represented in fig. 120(6) as far 
as known. The force polygon may be drawn thus: DC to 
represent 3000 lbs., CF to represent 1000 lbs., a line from F 
parallel to one of the unknowns and one from D parallel to the 
other. The last two lines determine E, and the force polygon 




a a 

Scale of force polygons: 1 in. = 2000 lbs. 
Fig. i 20. 



is DCFED ; hence the forces in the members de and ef are repre- 
sented by ED and FE (866 and 2500 lbs.) respectively. Both 
members are in compression. 

We next imagine the truss separated into parts as by 777. 
Choosing the part within III , we have a simple system to deal 
with ; the forces of it are represented as far as known in fig. 1 20(c) . 
Their force polygon may be drawn thus: AD to represent 2600 
lbs., DE to represent 866 lbs., a line from E parallel to one of 
the unknowns, and a line from A parallel to the other. The 
last two lines determine G, and the force polygon is ADEGA ; 
hence the forces in the members eg and ag are represented by 



§ VII.] JOINTED FRAMES. 145 

EG and GA (866 and 1734 lbs.) respectively. Each member 
is in tension. 

On account of the symmetry of the truss and loading, the 
forces in the remaining members are now known. 

150. Polygon for a Joint. — In drawing the force polygon 
for all the external forces on the part of a truss included within 
a small circle struck from a joint, it will be advantageous to 
represent the forces in the order in which they occur about the 
joint. 

A force polygon so drawn will be called a polygon for the 
joint; and for brevity, if the order taken is clockwise, the poly- 
gon will be called a clockwise polygon, and if counter-clockwise 
it will be called a counter-clockwise polygon. ABC DA (fig. 
1 20a') is a clockwise polygon for joint / of fig. 119; ABCD'A 
is a force polygon for the "forces at joint /," but it is not a 
polygon for the joint, because the forces are not represented in 
the polygon in the order in which the forces occur about the joint. 

The student should draw the counter-clockwise polygon 
for the joint and compare with ABCDA. 

151. Stress Diagrams. — If the polygons for all the joints of 
a truss are drawn separately as in the illustration in art. 149, 
then the stress in each member will have 
been represented twice. It is possible 
to combine the polygons so that it will 
not be necessary to represent the stress 
in any member more than once, thus 
reducing the number of lines to be 
drawn. Such a combination of force 
polygons is called a stress diagram. 

Figure 121 is a stress diagram for 
the truss of fig. 119 loaded as there 
shown. Comparing the part consisting v \ j 

of solid lines with figs. 120 (a'), (b f ), 
and (c r ), it is seen to be a combination j 

. iB' 

of the latter three figures. It will also p IG I2I 

be observed that the polygons are all 

clockwise polygons, but counter-clockwise polygons could be 

combined into a stress diagram. 




146 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VL 

To construct a stress diagram for a truss under given loads: 
(i) Letter the truss diagram as directed in art. 149; 

(2) Determine the reactions; 

(3) Construct a force polygon for all the external forces applied 

to the truss (loads and reactions), representing them in 
the order in which their application points occur about 
the truss, clockwise or counter-clockwise.* 

(4) On the sides of that polygon, construct the polygons for 

all the joints. They must be clockwise or counter-clock- 
wise ones according as the polygon for the loads and 
reactions was drawn clockwise or counter-clockwise. 
(The first polygon drawn must be for a joint at which but 
two members are fastened; the joints at the supports are 
usually such. Next that joint is considered (and its poly- 
gon is drawn) at which not more than two stresses are 
unknown, that is, of all the members fastened at that 
joint the forces in not more than two are unknown. Then 
the next joint at which not more than two stresses are 
unknown is considered; etc.f) 

These directions are illustrated in the following solutions. 

EXAMPLES. 

1. Solve ex. 4 of art. 147 by the graphical method. 

Solution: Supposing the reactions to have been deter- 
mined, we draw the force polygon for the loads and reactions 
ABCDEFA (fig. 1226); it is a clockwise polygon. We may 
begin by drawing the clockwise polygon for joint I or II; for the 
former it is FABGF.% Member bg is therefore in compression 



* The part of that polygon representing the loads is called a load line. 

f In some trusses, after the polygons for a few joints are drawn, there 
remains no joint at which there are but two unknown stresses; fig. 123 
represents such a one. The solution of ex. 5 explains several ways of 
procedure in such cases. 

% The student is urged to make sketches of the bodies (parts of truss) 
upon which the forces, whose polygons are being drawn, act. A force 
acting upon the "cut" end of a member and toward the joint is a push, 
and the stress in the member is compressive; if it acts away from the 
joint, it is a pull and the stress is tensile. 






§ VII.] 



JOINTED FRAMES. 



*47 



and gf in tension. Next we may draw the clockwise polygon 
for joint II, III, or IV; for the first it is CDEHC. Member ch 




(b) 



lin.= 4000 "lbs. 



lin.= 4000 lbs. 



Fig. 122. 



is in compression and eh in tension. For joint III, the polygon 
is HEFGH and member gh is in tension. If the work has been 
correctly and accurately done, the line GH is parallel to gh. 

2. Solve ex. 2 of art. 147 graphically. 

3. Solve ex. 3 of art. 147 graphically. 

4. Solve ex. 5 of art. 147 graphically. 

5. Analyze the truss of fig. 123 under the loads shown. 
Solution : Evidently each reaction equals one-half the whole 

load. ABCDEE'D'C'B'A'FA is a clockwise polygon for the 
loads and reactions. The polygon for joint 1 may be drawn 
first; it is FABGF. Next, that for joint 2 may be drawn; 
it is GBCHG. Then that for joint 3 may be drawn; it is 
FGHIF. The polygons for joints 1', 2' , and 3' are FA'B'G'F, 
G'B'C'H'G', and FG'H'I'F respectively. 

No joint remains at which there are but two unknown forces, 
and no more polygons can be drawn. If in any way the number 
of unknown forces at a joint can be reduced to two, the polygon 
for that joint can be drawn and the stress diagram can be com- 
pleted. There are several ways of making that reduction. 
For example, if the force in ij, jm, or mf were known, the poly- 
gon for joint 4 could be drawn, then that for 5, 6, 7, and 8. 

The force in mf may be determined by " passing a section" 
as at / and solving the external system on either part of the 
truss for the desired force. The system consists of the loads 
and reaction on that part and the forces in the members cut. 



148 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

The system may of course be solved graphically or algebrai- 
cally, but in this truss the algebraic solution is much the sim- 
pler. A moment equation for either system with joint 8 as 




Scale: 1 in.= 5000 lbs. 
Fig. 123. 

origin furnishes the value of the force in mf readily; the value 
is 34200 lbs. and the stress is tensile. 

This force in fm may now be represented in the proper place 
in the stress diagram determining M, and then the polygon for 
joint 4 can be drawn; it is MFIJM. The student should pick 
out the polygons for the remaining joints and determine the 
kind of stress in each member. 

There are other ways of meeting the difficulty presented in 
this form of truss, but that here given is the most general 
and can be applied readily to other forms. 

6. Analyze the truss represented in fig. 114(a) under the 
loads shown. 



§ VIIL] 



ROUGH SUPPORTS; FRICTION. 



149 




Fig. 124. 



§ VIIL Rough Supports; Friction. 

152. Definitions. — It is a fact of experience that when one 
body slides or tends to slide over another, the slid- 
ing of the first is opposed or resisted by the second. 
Thus, suppose that fig. 124 represents a block 
which slides or tends to slide over another body 
towards the right; the second body exerts some 
such force as R upon the block. 

The force which one body exerts upon another which slides 
or tends to slide over the first is called the total resistance; it 
will be denoted by R. The component of the total resistance 
along the (plane) surface of contact is called sliding resistance, 
or more commonly friction; the component normal to the 
surface is called normal pressure. (If the surface of contact 
of the two bodies is not plane, the force exerted at each ele- 
mentary part of the surface is the total resistance applied to 
that element, and its components in and normal to the element 
are the friction and the normal pressure applied to the element.) 

Friction is called kinetic or static according as sliding does 
or does not take place. Static friction only is here considered 
(kinetic friction is discussed later). 

Suppose that the block represented in fig. 125 weighs 10 lbs., 




that it is subjected to a horizontal pull, P, and the rubbing 
surfaces are such that P must exceed 6 lbs. to start the block. 
Fig. 125(a), (6), and (c) represent the forces acting upon the 
block when P, as it increases, reaches values of 2,4, and 6 lbs. 
respectively. Since the block is at rest, the friction at the three' 
stages equals 2, 4, and 6 lbs. and in all stages A T equals 10 lbs: 
When P reaches 6+ lbs. the block will move and the kinetic 



150 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

friction would be something less than 6 lbs., as has been dis- 
covered experimentally. For any two bodies then, the fric- 
tion may have many values depending on whether slipping 
occurs or not, and if not, on how great the tendency to 
slipping is. 

The friction corresponding to impending motion is called 
limiting friction; it will be denoted by F f . Evidently limiting 
friction is the maximum value of the friction corresponding to 
any given normal pressure, see fig. 125 (a), (b), and (c). Limit- 
ing friction has been studied experimentally and many impor- 
tant results have been thus deduced. 

153. The Coefficient of Static Friction for two rubbing sur- 
faces is the ratio of any normal pressure between the surfaces 
and the corresponding limiting friction. If it is denoted by /, 

f = F'/N, or F'=fN. 

154. The Angle of Friction for two rubbing surfaces is the 
angle between the directions of the normal pressure and the 
total resistance when motion is impending. Denoting it by 
<j> (see fig. 125c), 

tan <f> = F' '/N; hence tan <£ = /. 

I55» Angle of Repose. — If a block be placed upon an in- 
clined plane, the inclination at which slipping would be impend- 
JEJjflgflli ing is called the angle of repose for the two rub- 
l bing surfaces; it will be denoted by a. From 

Ob 




11 fig. 126 (representing a body on an incline, the 

w | \ N angle being that of repose), and the equations of 
Fig. 126. equilibrium for the forces, 

F' = W sin a and N = W cos a ; 

nence tan a = F'/N. 

Since F'/iV = /=tan <£, a = <f>, and tana = /; 

that is, the angle of repose for two surfaces equals their angle of 
friction, and the tangent of the angle of repose equals the coeffi- 
cient of friction. 



§VIIL] ROUGH SUPPORTS; FRICTION. 15* 

156. Laws of Friction. — The following laws relate to "solid 
friction ' ' (friction between solids) and are based entirely on 
experiment. 

1. The coefficient of friction for two surfaces depends upon 
the nature of the surfaces. Thus the coefficient varies with 
the materials, with the smoothness of the surfaces, and with 
the lubricant, if any is used. 

2. The coefficient of friction is independent of the normal 
pressure between the two surfaces and of the extent of the 
contact. This law is not exactly true; especially for such low 
pressures at which a considerable part of the sliding resistance 
is due to adhesion, and for high pressures which result in a 
change in the character of the surfaces; also when lubrication 
is excessive, for then the friction is mixed, being neither "solid" 
nor "fluid. " 

157. Determination and Values of the Coefficient. — The 
coefficient of friction for two surfaces may be determined by 
measuring their angle of repose (see fig. 126). The tangent of 
the angle is the coefficient sought. Or, the two bodies may 
be placed as in fig. 125, and then measuring the pull P neces- 
sary to start the block, F f is known since F f equals that pull. 
Also, N = W; hence f = P/W. 

In one of these two ways many determinations have been 
made, the values of the coefficient for a few materials ranging 
as follows: 

Wood on wood, soaped 0.22-0. 44 

" " " dry 0.30-0.70. 

Metal on metal, dry 0.15-0. 24 

" as in polished and well-lubricated 

bearings o . 05 - o . 08 

Wood on metal, dry o . 60 

Hemp rope on wood o . 50 - o . 80 

Sole-leather on wood or cast iron, as in packings, dry o . 40 - o . 60 

Leather belting on pulleys 0.25-1. 00 

Stone on stone, as in arches o . 40 - o . 60 




152 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

EXAMPLES. 

i. If the block represented in fig. 127 weighs 100 lbs., 6 is 

io°, and the coefficient of friction is 0.2, how 

great must P be to start the block? 

Solution: Suppose motion to be impending, 
Fig. 127. ,, ^ ^ s ' 

then 

P cos d = F' = o.2N = o. 2(100— P sin io°), 
or P= 19.62 lbs. 

A force slightly greater than this will start the block. 

2 . Solve ex. 1 if the sense of P is reversed. 

Ans. ,21.05 lbs. 

3. If Pin ex. 2 is 15 lbs., how great is the friction? 

4. What is the least value of P acting as shown in fig. 127 
that will start the block? What is the corresponding value 
of 0? Ans. d = ii° i8'.6. 

5. If the block represented in fig. 128 weighs 
100 lbs., /?=25°,/ = J, and d = o, how great must P 
be to start the block? , Ans. 72.46+ lbs. 

6. To prevent slipping, how great must P. be? 

7. If Pis 20 lbs., determine the friction. 

Ans. 22.26 lbs. 

8. If P is 50 lbs., determine the friction. 

9. Take 0= io° in ex. 5, and solve. Ans. 69.22 + lbs. 

10. Show that P (fig. 128) to start the body up is a mini- 
mum if d =tan -1 /- 

11. Two bodies connected by a cord are placed upon a 
plane inclined 12 to the horizontal, the string being taut but 
without initial tension and inclined 12 with the level base. 
If the bodies weigh 10 and 15 lbs. and the corresponding coeffi- 
cients of friction are \ and J, determine the frictions. 

12. In the preceding example, change the coefficient -J to 
\ and suppose the lower body to be the lighter one. Solve and 
also determine the tension. 

13. A bar weighing 100 lbs. rests upon two end supports at 
the same level. Suppose a force is applied to it so that the 
action line passes through the points of support. If the coeffi- 
cients of friction for the rubbing surfaces be 0.2 and 0.25, how 




§ VIII.] 



ROUGH SUPPORTS: FRICTION. 



153 



great must the force be to move the bar? What can you say 
about the frictions when the force is 18 lbs. and when it is 
20 lbs.? 

14. How great must P (fig. 129) be to start the wedge 
against the force Q? 

(b) 1 

Y M I 

^'' R"l£ \ 




Solution: When the wedge is about to slip, the directions 
of the resistances at the rubbing surfaces are known, for the 
inclination of each to the normal to the surface on which it acts 
equals the corresponding angle of friction. Of the three forces, 
Q, R', and R" , applied upon the upper body, the direction of all 
and the magnitude of one, Q, are known. Their force triangle 
determines the magnitude and sense of R' and R" (see fig. 1296). 
Having determined R", P and R"' may be determined by means 
of the force triangle for those forces (see fig. 129c). 

15. How great must P be to prevent the wedge from 
slipping out, the wedge angle being 25 , §" and $'" io°? 

16. Show that the wedge will not slip out \ 
if its angle is less than <£"-}- <£'" when P=o. 

17. Fig. 130 represents a jack-screw. How 
great a couple whose plane is horizontal must 
be applied to the screw to " overcome " Q? 

Solution: At each point of the lower sur- 
faces of the thread on the screw, the nut 
exerts a pressure whose normal and tan- 
gential components call dN and dF respect- 
ively. When the tendency of the screw is FlG ' I3 °* 
to rise, dF acts downward. Let C denote the moment of the 



4-Q 



1 n 


1 


"x 


s. 


_dN 




154 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



couple, /? the pitch angle, and r the average arm of the frictions 
and normal pressures with respect to the screw axis. From 
the conditions of equilibrium (see art. 125), 

ZFy = Q - 2(dF sin /?) + Z(dN cos /?) = o ; 
IM y = C-I(dF cos/9-r) -2(dN sin /?-r)=o; 

and, when slipping is impending, dF = fdN. 
These three equations make 

C = Qr(sin fi + f cos /3)/(cos /?-/ sin /?). 

18. Show that to cause the screw to descend, 

C = Qr(f cos /?-sin /?)/(/ sin /? + cos /?). 

19. Show that if /?><£, the screw will descend under the 
action of Q alone. 

20. Fig, 131 represents a lever supported in a triangular 
bearing. How great a force, P, is required to " overcome" Q? 

Solution: When slipping is about to occur, the reactions 
at A and B act in the directions indicated. Since the action 
lines of P, Q, R', and R" and the magnitude of Q are known, 
the magnitude of P (and of R' and R") may be determined (see 
art. 141). 





Fig. 131. 

158. Friction Circle. — Fig. 132 represents a journal and its 
bearing (also a pin joint), the fit being loose so that the con- 
tact is along a line practically. Let r denote the radius of the 
journal and <j> the angle of friction for the rubbing surfaces. 
Then a circle concentric with a cross-section of the journal 
whose radius equals r sin <j) is called the friction circle for the 
journal and bearing. It is useful in solving certain problems 
involving the friction of a loose-fitting journal and bearing, pin 
joint, etc. 



5 VIII.] ROUGH SUPPORTS; FRICTION. *55 

Proposition. — When slipping is about to occur at a loose 
bearing, the action line of the resistance offered by the bearing 
is tangent to the friction circle. 

Proof: The resistance is applied at the line of contact, rep- 
resented by A, and it makes an angle <f> with the normal AC 
(art. 154). If AB is tangent to the circle, then 

sin BAC = r sin <£/r = sin <£, or BAC = <f)\ 

hence the tangent line and the action line of the resistance 
coincide. 

Remembering that Hie tangential component of the re- 
sistance (the friction) opposes the tendency to slip, the student 
will have no difficulty to tell which one of the two tangent lines 
which may be drawn is the action line of the resistance. 

EXAMPLES. 

1. Fig. 133 represents a lever supported in a loose cylindri- 
cal bearing. How great a force --Vc 

Pis required to "overcome" Q? / \ Z^~~s^ 

Solution: There being three yf^^/^\^^^ 
forces applied to the lever P, Q, *v. ^^1 /'TV C_^> 
and the resistance of the bearing ^^^~"'jM 

R, their action lines intersect in ^mftm^ 

a point (art. 125); hence R passes IG ' I33 ' 

through D. Since R is also tangent to the friction circle, its 
action line is determined, and the system P,Q,R may be solved 
for the unknown magnitudes. 

2. Let MCN (fig. 133) equal 90 , CM 2 ft., CN 6 in., the 
angles CND and CMD 6o° and 90 respectively, radius of 
axle 2 in., the coefficient of friction J, and Q 1000 lbs. Deter- 
mine P and the friction. 

3. Determine the largest force P which Q will overcome, 
data as in ex. 2. 

4. When the lever is straight (7VCM=i8o°) and P and Q 
act at right angles to NCM, show that 

P(CM^r sin (j>)=Q(CN±r sin <£), 

according as the impending motion is with P or Q. 



156 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 




Fig. 



134. 



159. Cone of Friction. — Let P (fig. 134) denote the result- 
ant of all the forces applied to the body repre- 
sented, not including the resistance of the sup- 
porting surface. Then the cone whose apex is 
at C, whose axis is the normal through C y and 
whose apex angle equals twice the angle of friction 
is called a cone of friction. In the figure the cone 
is represented by ACB. 
Proposition. — If the action line of the resultant of all the 
forces applied to a body not including the resistance of the 
support falls within the cone of friction, sliding will not occur; 
if without, it will occur. 

Proof: The force causing or tending to cause sliding is the 
horizontal component of P (fig. 134), equal to P sin 6. Since 
the maximum sliding resistance, or friction, is 
F'=tan 0.JV = tan <f>-P cos 6, 
(sliding force)/F' = tan 0/tan <£. Therefore 
if d>cj) (P falls without the cone), 

the sliding force > limiting friction ; 
if 6<<f> (P falls within the cone), 

the sliding force < limiting friction. 

EXAMPLES. 

1. In fig. 128 suppose that 6 and P — 30 , P = $o lbs.,/ = J, 
and the body on the incline weighs 100 lbs. Determine whether 
the body will slide and the value of the friction graphically. 

2. Let P in the preceding example be 10 lbs., and solve. 

3. A prismatic block of wood is sawed into two parts so that 
the cut is inclined at an angle with the axis. If the two parts 
are laid together matching and end pushes are applied along 
the axis, for which values of 6 will slipping not occur? 

4. Fig. 135 represents a slider which may slide in the guides 
A and B; </> f and <£" are the angles of 
friction for the rubbing surfaces respect- 
ively. Show that any horizontal force 
applied above C cannot move the slider. 

Solution : Let P be a horizontal force 
applied above C, and imagine it resolved 
into two components, P' and P n ', whose 



a 

W//A 


h' ajr 


1 P 


A 


S 


/ 


Vfgv 


< 


,A 


WM%%%. 



Fig. 135. 



§ VIII.. 



ROUGH SUPPORTS; FRICTION. 



i5' 



action line passes through A and B respectively. P' and p" 
may be regarded as the forces overcoming the resistances at A 
and B. Now it is impossible to resolve so that P' and P" will 
fall without the respective cones of friction and tend to move 
the slide in the same direction. For, if P is resolved at any 
point to the left of a, P' and P" tend to move the slider to the 
left and right respectively; and if it is resolved at any point 
to the right of b, P' and P" urge the slider to the right and left 
respectively. If P is resolved at any point between a and 6, 
both Components do not fall without the respective cones of 
friction; hence slipping cannot occur. 

Or thus: imagine slipping about to occur at B; then the 
resistance there acts along the line BC . To keep the slider at 
rest, the resistance at A must act through b f , and since the line 
Ab f is within the cone at A, such resistance is possible, and 
equilibrium will be preserved. 

5. Fig. 136 represents a bar AB resting in 
position upon two inclined planes, 
<f>' and <j>" being the angles of fric- 
tion at A and B respectively. Show 
that if the weight of the bar is neg- 
lected, any body Suspended from a 
point between m and n will not cause 
it to slide, but that if suspended 
beyond m or nit will cause sliding. 

(a) ( c ) 6. The front of a drawer is 4 ft., 

r and its sides 10 in. long. If the 
angle of friction for the rubbing 
surfaces at the sides is 22 , and the 
drawer handles are 3 ft. apart, show 
that the drawer cannot be opened 
by a forward pull at one handle. 

160. Belt Friction. — Fig. 137(a) 
represents a cylinder about a part 
of which a belt (or cord) is wrapped. 
If the cylinder is not smooth, the 
pulls T 1 and T 2 may be quite un- 
equal without causing slipping of 
the belt, as may easily be verified by trial. 





158 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



The forces acting upon the part of the belt in contact with 
the cylinder consist of the tensions T x and T 2 , the normal 
pressure and the friction (see fig. 1376). Let p denote the nor- 
mal pressure per unit length of arc ; then the normal pressure 
on any part whose length is ds (fig. 137c) is pds. The friction 
on that part may be called dF and the tensions T and T + dT. 
Since the portion is in equilibrium, 



~ . dd 
2 1 sin — 
2 



Tdd: 



hence 



p-T/r- (1) 

that is, the normal pressure per unit length at any point of the 
contact equals the belt tension there divided by the radius of 
the cylinder. 

When slipping is impending, dF = f-pds, and since dF = dT, 

dT 

T - T 'H 



dT 



Integration gives 



T 
ffds, 



or 



f d i=m 



hence 



or 



log, 
log, T 2 



T 



log. 



(see fig. 137a); 



w. 



T 2 = T^. 



(2) 
(3) 



The angle /? must be expressed in radians; e is the base of the 
Naperian system of logarithms, 2.718. The formulas apply 
also when /? is greater than 277:, that is, when the cord more than 
encircles the cylinder. 

For a given value of T lf T 2 increases very rapidly with /? 
as shown by fig. 138 which represents the locus of equation (3), 
T 2 and ft are the variables and e, f, and 
T x constants, / being taken as ^ and 
T x = OA. To the scale OA = T lt OB rep- 
resents T 2 when ft = AOB. 

EXAMPLES. 

1. Compute the ratio between 7\ and 
T 2 , when / is J and the cord is wrapped 
twice around the cylinder. 
2. Plot in fig. 138 the locus of equation (3) when f is \ 




Fig. 138. 



§K.] 



FORCES IN SPACE AND MISCELLANEOUS. 



59 



§ IX. Forces in Space and Miscellaneous. 

161. Examples Involving Non-Parallel Non-Coplanar Forces. 

— The principles for solving such examples are stated in art. 125. 
Sometimes in simple cases the example can be resolved into 
others the forces in which are coplanar. Such separation is 
usually a simplification; ex. 3 is an illustration. 

EXAMPLES. 

i . Determine the relation between P and W of the windlass 
represented in fig. 139 and the reactions of the bearings in terms 




Fig. 139. 

of P. Take P always at right angles to the crank as shown, 
and neglect friction. 

Solution: The forces acting on the windlass are P, W, and 
the reactions of the bearings, A and B\ the weight of the wind- 
lass is neglected. Let R' and R" denote the reactions at 
^4 and B respectively, and a the angle which the crank makes 
with the x axis; then 

IF X = -P sin a +R£+R£' = o, 

IF y = P cos a +Ri+R^-W=o y 
1M X = -Pcosa-b-Wx+R{,a = o 1 



2M, 



P sina-b— R x ' a = o, 



IM Z = Pc-Wr = o, 



r being the distance from the axis of the rope to that of the 
windlass. 

The ^-resolution equation vanishes since none of the forces 
have z components. Solving the equations, we find that 



160 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

W = Pc/r, 
R' x = — P sin a -b/a, R[ = P(b cos a +cx/r) /a, 

R%= P sin a(i +b/a), Ri' = P(i ~x/a)c/r-P cos a(i +6/a). 

2. Fig. 140(a) represents a derrick consisting of a post (7L£), 
boom (AC), two "stiff legs" (Z?L> and BE), and hoisting cables 
which are not shown in detail. Determine the forces on the 
parts of the derrick due to the load W, assuming for simplicity 
that B and C are merely connected by a cable, that the load is 




1 (6) 

Pig. 140. 

suspended from C, and that the boom is connected to the post 
at its lower end practically. 

Solution: The external forces on the whole derrick are W 
and the reactions at A, D, and E. The directions of these reac- 
tions will depend on the nature of the supports at A, D, and E; 
we will assume that these are such that the reaction at A is a 
single force acting through A, and that those at D and E act 
along DB and EB respectively. 

We first resolve this system into two component systems, 
one coplanar (its plane being that of the ground) and one par- 
allel (its forces being vertical). Let the reactions at D and E 
be called R' and R" respectively; then the components of the 
R' are 

R' cos 45 acting in the line AD, 
R' sin 45 " vertically at D, 

and the components of R" are 

R" cos 45 acting in the line AE, 



R" sin 45 



vertically at E. 



§ix.] 



FORCES IN SPACE AND MISCELLANEOUS. 



161 



Let the reaction at A be called R'", and imagine it resolved into 
x, y, and z components at A. The coplanar component system 
consists of forces as shown in fig. 140(6), and the parallel com- 
ponent system consists of four vertical forces, R f sin 45 at D, 
R" sin 45 at E, R*/' at A, and W at C (all not shown). 

Each of the component systems being in equilibrium, we 
may write the appropriate equations of equilibrium for each; 
thus for the first (concurrent) 

IF X = -R f cos 45 cos 45°-^" cos 45 cos 4 5°+i^" = o, 
IF Z = -R' cos 45 sin 4$°+ R " cos 45° sin 45°+^" = 0; 
and for the second 



2Fy=Rl" 



sin 45 = 0, 



W-R' sin 45° -R 

IM x =Wb cos <j> smd—R f sin 45°- a cos 45 



+ R" sin 45 -a cos 45° = o, 
JM Z = — Wb cos 0cos d+R' sin 45 -a cos 45 

+ R" sin 45 -a cos 45° = o. 

These equations determine the five unknown forces R' , R", 
R£", Rl", and R' z " . The solution is left to the student. 

The tension in the cable BC and the pressure between the 
post and the boom may be determined from a consideration of 
the forces acting on the boom. As these forces are coplanar 
their determination is left to the student. 

3. Fig. 141(a) represents a "shear-legs crane." It consists 
of two posts, CD and CE, hinged together at the top and hinged 






I 

(*) 

Fig. 141. 
at their bases so that they can rotate about the line joining D 
and E. The stay AC may be a cable or a "stiff leg"; if stiff, 
AC is constant in length and the load is swung in or out (the 



1 62 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 

posts rotating about DE) by moving the lower end of A along the 
track AB. Determine the forces on the parts due to the load W. 
Solution: Imagine that the two posts are replaced by a sin- 
gle one CB, and determine the tension in the stay and the com- 
pression (R) in CB from a consideration of the forces applied 
(see fig. 1 41 b). Then resolve this compression R into two 
components whose action lines coincide with the axes of the 
posts (see fig. 141c). 

4. In fig. 141 take AC = $o feet, BD = BE=io feet, = 6o°, 
W=io tons, and solve the preceding example graphically. 

5. Fig. 142 represents a small dipper dredge, the side eleva- 
tion (a) representing a position for filling the dipper, and the 




Scale, 1 in =20 ft. 

Fig. 142. 



(6) 



front elevation (b) a position for emptying. The boom swings 
about a vertical axis at its lower end, a ratchet on the lower 
side of the dipper handle engages a pinion on the boom by 
means of which the effective length of the handle can be changed, 
and the two back stays BC are fastened at points 10 feet apart, 
36 feet to the left of D. In (a) assume that the dipper is stuck, 
the pull in the chain being 5000 lbs., and in (b) that the dipper 
load is 1000 lbs. Neglect the weights of all parts and deter- 
mine in each case the tensions in the stays AB and BC, the 
compressions in the posts of the "A frame," the reaction at 
the pivot at the base of the boom, and the pressure on the 
ratchet. Solve graphically. 

6. Fig. 143(a) represents a giant wharf-crane. The struc- 
ture consists of a rigid framework, T-shaped, and a rigid tripod. 



§IX.] 



FORCES IN SPACE AND MISCELLANEOUS. 



163 



The stem of the T stands within the tripod and rests against 
the tripod head and on rollers at the base. Determine the reac- 
tions at the base of the tripod legs, as far as possible, due to a 
load P of 150 tons at the position shown in the figure. When 
the cross-piece of the T is represented in plan by A a, the short 
arm being on the side A , what are the reactions ? 




^v/^yM/w. ^ r^^ ^ /ww/rVW /W^''''"^^ 



Fig. 143. 

7 . The crane of the preceding example is revolved by means 
of a circular rack on the tripod head and two pinions at oppo- 
site ends of a diameter of the rack. It is estimated that when 
the crane is being turned against a maximum resistance (due 
to friction, inertia, and wind-pressure), the reaction on the rack 
is a horizontal couple of 112,000 foot-pounds. The diameter 
of the rack being 22 feet compute the reactions at the base of 
the tripod legs due to the reaction mentioned. 

162. Miscellaneous. — The principles for the solution of the 
following examples are given in art. 125. Similar examples 
have been worked in the first part of this chapter, and the stu- 
dent should have no difficulty in solving the following set. 

EXAMPLES. 

1. Fig. 144(a) represents a platform scale which consists 
of a frame FF with five knife-edges, one at K v two at K 2 , and 



1 64 APPLICATIONS OF THE PRINCIPLES OF EQUILIBRIUM. [Ch. VI. 



two at K s . The platform rests on tfyree levers which bear upon 
the four knife-edges at K 2 and K 3 . The short levers are also sup- 
ported by the long one at 5, and the long one is connected by 




\\n in 



V'M^/// 




l< — G-^Ma 



Fig. 144(a). 



a vertical rod to the "scale-beam" PK V Determine the rela- 
tion between the weight of the poise (P) and that of the body 
(W) on the platform, and show that it is independent of the 
position of the body. 

2. Assume that the steam-pressure P just balances the load 
W on the hoisting-engine (fig. 1446). If P=iooo lbs. and the 
angle BCA is 6o°, compute the compression in the connecting- 
rod, the pressure against the cross-head guide, the tangential 
component of the crank-pin pressure, and W. 





wo w 

Fig 144(6). Fig. 144(c). 

3. Fig. 144(c) represents a "crab hook." The lengths A B 
and BC are 12 and 21 im respectively, the angle ABC is ioo°, 
CD = 12", and BB' = ^ ft. Determine the stresses in CD and 



§IX.] 



FORCES IN SPACE AND MISCELLANEOUS. 



165 



BB' if W=iooo lbs. and AA' = i2>", the weight of the parts 
being neglected. 

4. Fig. 144(d) represents a simple elevator-car. What are 
the pressures of the wheels on the rails due to the load W ? 





W/W///////////////7////////// 

Fig. 144O?). 

5. Fig. i44(^) represents a hand-press, consisting of a lever 
ACP and two short links BC pinned to a cross-head at their 
lower ends; AC = BC=i$". Compute the pressure on the bale 
in the press and the pin-pressures when AB = 2 ft., and the arm 
of P with respect to A is 5 ft., P being 100 lbs. 



KINEMATICS. 



CHAPTER VII. 
RECTILINEAR MOTION OF A PARTICLE. 

§ I. Velocity and Acceleration. 

163. Specification of Position. — The position of a point in a 
given line can be specified by a single quantity, namely, the 
abscissa of the point with respect to any other point in the line 
assumed as "origin." Thus if the abscissas of points to the 
right of the origin O (fig. 145) be given H , 

the plus sign and those of points to the — ■ *— 

left the minus sign, then P' is specified FlG - I4 5- 

by the abscissa § inch and P" by — \ inch. Position abscissas 
will be denoted by x. 

164. Space-Time Curves. — A rectilinear motion of a point 
can be well represented by means of a line called the space-time 
curve for the motion. This is a line the ordinate and abscissa 
to any point of which represent the position abscissa and corre- 

sponding value of the time respect- 

y\ ]\ ively. To construct this curve plot 

r j^ 1 1 \ T corresponding values of x and the time 
i T -l \ along vertical and horizontal axes as 

*r "* ^ shown in fig. 146, and join all such 

plotted points; the connecting line is 
FlG - x 4 6 - the space-time curve. 

Evidently the space-time curve for a motion gives the posi- 
tion of the moving point at each instant, and it is therefore a 
complete record of the motion. 

165. Displacement. — If x t denotes the abscissa of a moving 
point at the instant t v and x 2 that at a later instant t 2 , then the 

167 



1 68 



RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 



displacement for the interval t 2 — t x is denned as x 2 —x v Evi- 
dently a displacement (x 2 — x 1 ) may be positive or negative; 
hence, two displacements must agree in sign as well as in mag- 
nitude to be equal. 

166. Kinds of Rectilinear Motion. — If the displacements of 
a point in equal intervals of time (large or small) are equal, the 
motion is called uniform; and if the displacements are not equal, 
the motion is called non-uniform. Non-uniform motions are 
further classified as explained in art. 172. 

The space-time curve for a uniform motion is obviously an 
inclined straight line, and the space-time curve for a non-uni- 
form motion is a curved line. 

QUESTIONS. 

i. What is the difference between the motions represented 
in fig. 147 (a) and (&)? 

2. What can you tell of the motion whose space-time curve 
is that in fig. 147(c)? 

3. Are horizontal and vertical space-time curves possible? 




Fig. 147. 

167. Velocity. — The velocity of a moving point is the rate 
with respect to time at which it changes position, or at which 
its displacement occurs. Still otherwise stated, it is the time- 
rate of (change of) the position abscissa of the moving point. 

Let x and t denote position abscissa and time respectively; 
then, as shown in works on calculus (and in Appendix B), the 
time-rate of x is dx/dt; hence, if v denotes velocity, 

v = dx/dt (1) 

This gives the value of v at any instant / ; its value at a particu- 
lar instant equals the value of dx/dt for that instant. 

// the motion is uniform, x changes uniformly, and the time- 



§ I.] VELOCITY AND ACCELERATION. 169 

rate of x is Ax/ At, Ax denoting the displacement which occurs 
in any interval At. Hence 

v = Ax/At, (2) 

and plainly the velocity is constant. 

Since Ax/ At is the average time-rate of the displacement, 
equation (2) gives also the average velocity when applied to 
non-uniform motions 

Since the space-time curve is the "locus" or "graph" of the 
equation between x and t, dx/dt is the general expression for the 
slope or gradient of that curve. Hence the velocity correspond- 
ing to any point on a space-time curve is represented by the 
slope of the curve at that point. We say "is represented by" 
instead of equals, because, while the velocity at a certain instant 
is definite, the slope depends on the scale used in plotting tlje 
space-time curve. The slopes must therefore be interpreted by 
scale or be computed in a certain way, as explained in ex. 1, 
art. 169. 

168. Unit Velocity. — The expressions for velocity in eqs. (1) 
and (2) of the preceding article imply a certain unit of velocity, 
namely, the velocity of a point moving uniformly and so that 
it describes unit distance in unit time. Specific units of velocity 
are one foot-per-second, one mile-per-hour, etc. There are no 
short names for these units except for the nautical mile-per- 
hour, which is called a knot. 

The term per is conveniently replaced by the solidus , / ; foot- 
per-second, mile-per-hour, etc., are abbreviated thus: ft. /sec, 
mi./hr., etc.* 

169. Sign of a Velocity. — The expressions dx/dt and Ax/ At 
may be positive or negative; therefore v must be regarded as 
having the same sign as that of dx/dt or Ax/ At (see eqs. (1) and 
(2), art. 167). When the point is moving in the positive direc- 
tion, dx/dt and Ax/ At are positive, and when it is moving in the 
negative direction they are negative (see fig. 147); hence 

the sign of the velocity of a moving point at any instant is 
the same as that of the direction in which it is then moving. 

* For dimensions of a unit velocity see Appendix C. 



X 


/ 


'a 




/ 1 




T 




X scale: Tin.=60 fib. 




T scale: 1 in." 


=8 sec. 



170 RECTILINEAR MOTION OF A PARTICLE. [Chap. VIL 

EXAMPLES. 

i. What is the velocity when 2 = 2 sees, in the motion whose 
space-time curve is shown in rig. 148? 

Solution : We find first the point P of the curve correspond- 
ing to t = 2 sees., and then draw a tangent 
to the curve at that point. Next we take 
any point Q in the tangent line, and from 
it draw a perpendicular to the horizontal 
through P. Then we measure by scale 
the lines QR and PR, and take their ratio 
as measured. We find that QR = 3$ ft. 
and PR = 4 sees.; hence 

Fig. 148. t; = 35/4 = 8j ft.-per-sec. 

2. A point moves so that x = ct 3 , c being a constant. Show 
that its velocity at any time t is 3c* 2 . 

3. Let c in the preceding ex. be 10, x being in ft. and / in 
sees. When t = $ sees., where in its path is the moving point, 
and what is its velocity? Ans. ^ = 270 ft. /sec. 

4. A point moves so that # = 100/, x and / being in ft. and 
sees, respectively. What is its velocity? 

Solution: Here x varies uniformly; hence v = dx/Jt. From 
the law of the motion, Jx=iooJt, or v = 100 ft. /sec. (Can the 
value of v be deduced from eq. (1), art. 167?) 

5. A body falls in a vacuum according to the relation x = 
16. it 2 , x and t being in ft. and sees, respectively. What is the 
formula for the velocity ? 

6. Draw a space-time curve for the motion of a falling body. 

7. A point moves so that x= lot — t 2 , x and t being in ft. and 
sees, respectively. What is its velocity when t = 6 sees.? 

8. A point moves so that x = c cos (kt), c and k being con- 
stants. Deduce an expression for its velocity at any time t. 
Also let c = 2, k = 3, x and / being in ft. and sees, respectively, 
and compute the velocity when £ = 4 sees. 

9. A sprint of 100 yards being accomplished in 10 sees., what 
was the sprinter's average velocity in ft.-per-sec.? In mi.- 
per-hr. ? 



§ I.] VELOCITY AND ACCELERATION. I7 1 

10. In a certain motion v = ^t 2 , v and t being expressed in 
ft.-per-sec. and sees, respectively. Determine the displacement 
in the interval from the second to the fourth sec. 

Solution: Since v = $t 2 = dx/dt, 

dx = $t 2 dt, or x = t 3 + C, 
C being a constant of integration whose value depends on the 
mode of reckoning % and t, not specified.* Let x 2 and x 4 denote 
the values of x when t = 2 and 4 sees, respectively; then 

# 4 = 4 3 -fC = 64+C, and x 2 = 2 3 + C = 8 + C. 
Hence x 4 — x 2 = 64 — 8 = 56 ft. 

Instead of introducing a constant of integration we might 
integrate between limits; thus, from dx=7,t 2 dt, 

f* 4 dx = 3£t*dt, or x 4 -x 2 = 3 r^1 4 = 56ft. 

n. In a certain motion ^ = 3^ + 4, v and t being expressed 
in ft.-per-sec. and sees, respectively. If the moving particle is 
6 ft. to the right of the origin at the instant from which t is 
reckoned, determine the position at any time t and draw the 
space-time curve for the motion. 

170. Velocity-Time Curve. — The way in which the velocity of 
a moving point changes with respect to time can be represented 
graphically by a line called the velocity-time curve for the motion. 
This is a line the ordinate and abscissa of any point of which 
represent the velocity and the corresponding value of the time 
respectively. 

To construct this curve plot corresponding values of velocity 
(v) and time (t) along vertical and horizontal axes respect- 

* Note on the Determination of Constants of Integration. — The student 
is reminded that to determine a constant of integration he has only to 
substitute for the variables in an equation containing the constant any 
simultaneous values of them and then solve for the constant. Thus in 
the case above, suppose it had been stated that x is measured from the 
place occupied by the moving particle at the instant from which t is 
reckoned; then when t was zero x was also zero, i.e., simultaneous values 
of x and t are x=o and t=o. These substituted in the equation con- 
taining C make it 

o=o 3 + C; hence C =0. 



172 



RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 



ively, as shown in fig. 149, and join all such plotted points. 
The connecting line is the velocity-time curve. 

171. Velocity Increment. — If v x denotes the velocity of a 
point at an instant t lf and v 2 that at a 
later instant t 2 ,then v 2 — v t (not 2^— v 2 )is 
the velocity increment for the interval 
t 2 — t v Evidently a velocity increment 
maybe positive orjiegative; hence, two 
velocity increments must agree in sign 

FlG - I 49- as well as in magnitude to be equal. 

172. Kinds of Non-Uniform Motion. — A non-uniform motion 
whose velocity increments for equal intervals (large or small) 
are equal is called uniformly varying; one whose velocity incre- 
ments are unequal is called non-uniformly varying. 

Evidently the velocity-time curve for a uniformly varying 







(6) 

Fig. 150. 

motion is a straight line (fig. 150 a and b) and that for a non- 
uniformly varying one is a curved line (fig. 150c). 

QUESTIONS. 

1. What is the difference in the motions whose velocity- 
time curves are shown in fig. 150 (a) and (6)? 

2. What can you say of the motion whose velocity-time 
curve is shown in fig. 150(c)? 

3. Are horizontal or vertical velocity-time curves possible? 
173. Acceleration. — By acceleration of a moving point is 

meant the rate at which its velocity changes with respect to 
time, or simply the time-rate of (change of) its velocity. 

Let v and t denote velocity and time respectively; then, as 
shown in works on calculus (and in Appendix B), the time-rate 
of v is dv/dt\ hence if a denotes acceleration, 

a = dv/dt = d 2 x/dt 2 (1) 



§ I.J VELOCITY AND ACCELERATION. 173 

Equation (i) gives the value of a at any instant t; its value at 
a particular instant equals the value of dv/dt or d 2 x/dt 2 for that 
instant. 

// the motion is uniformly varying, v changes uniformly and 
the time-rate of v is Av/At, Av denoting the velocity increment 
for any interval At. Hence in this case 

& = Jv/Jt, (2) 

and plainly the acceleration is constant. 

Since Av/At is the average time-rate of the velocity, equa- 
tion (2) gives also the average acceleration when applied to 
non-uniform motions non-uniformly varying. 

Since the velocity-time curve is the locus or graph of the 
equation between v and t, dv/dt is the general expression for 
the slope or gradient of that curve. Hence the acceleration 
corresponding to any point on a velocity-time curve is repre- 
sented by the slope of the curve at that point. The slopes 
must be interpreted by a scale, or be computed in a certain way 
as explained in ex. 1, art. 175. 

174. Unit Acceleration. — The expressions for acceleration in 
eqs. (1) and (2) art. 173 imply a certain unit of acceleration, 
namely, the acceleration of a point whose velocity varies uni- 
formly and so that it changes by a unit in each unit time. 
Specific units of acceleration are 

one knot-per-hour (one nautical mile-per-hour-per-hour) 

one foot-per-second-per-second, etc. 
Abbreviating the term per as before, the above-named units are 
written thus: knot/hr. (mi./hr./hr.), ft. /sec. /sec. ; or still more 
briefly, mi./hr 2 , ft. /sec. 2 * 

175. Sign of an Acceleration. — The expressions dv/dt and 
Av/At may be positive or negative ; therefore a must be regarded 
as having the same sign as that of dv/dt or Av/At (see eqs. (1) 
and (2), art. 173). Now when the velocity increases alge- 
braically, dv/dt and Av/At are positive, and when it decreases 
algebraically, dv/dt and Av/At are negative (see fig. 150c); 
hence 

* For dimensions of a unit acceleration see Appendix C. 



174 RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 

the sign of the acceleration of a point at any in- 
stant is positive or negative according as the ve- 
locity is then increasing or decreasing (algebraically). 



EXAMPLES. 

i. What is the acceleration when / = 3 sees, in the motion 
whose velocity-time curve is represented in fig. 151? 

., Solution: First we find the point 

P on the curve corresponding to t = 3 
sees., and then draw a tangent to the 
curve at that point. Next we take 
any point Q on the tangent and draw 
from it a perpendicular to the hori- 



v scale: i in.=i6 ft. per sec. zontal line through P. Then we meas- 
Tscaie:iin.=8sec. ure by scale the lines QR and PR and 

FlG - I 5 I - take their ratio as so measured. We 

find that QR = 7 ft.-per-sec. and PR = 8 sees. ; hence 

a = 7/8 = 0.875 ft.-per-sec.-per-sec. 

2. In a certain experiment on "getting up speed" of electric 
trains the speeds recorded at 5 -minute intervals were as follows : 
o, 19, 30, 35, 38, and 40.5 mi.-per-hr. Draw the velocity- time 
curve for the motion and determine the accelerations at the 
beginning and end of the period. 

3. The motion of a point being according to the equation 
x = ct 3 , c being a constant, show that the acceleration equals 6ct. 

4. Let c in the preceding example be 10, x being in ft. and t 
in sees. What is the value of the acceleration when 2 = 3 sees.? 

5. A point moves so that v = $ot, v and t being in ft.-per-sec. 
and sees, respectively. What is its acceleration? 

Solution: Since v varies uniformly, a = Av/At. From the 
law of the motion Jv = $oJt; hence a = 50 ft.-per-sec. (Can the 
value of a be deduced from eq. (1), art. 173 ?) 

6. A body falls in a vacuum according to the law x= 16. it 2 , 
x and t being in feet and seconds respectively. What is the 
value of the acceleration? 



§L] VELOCITY AMD ACCELERATION. 175 

7 . Draw a velocity-time curve for a body falling in vacuum, 
and compare with the curve drawn in your solution of ex. 6, 
art. 169. 

8. A point moves so that x = iot— t 2 , x and t being in ft. 
and sees, respectively. What is its acceleration when t = 6 
sees. ? 

9. A point moves so that x = ccos (kt), c and k being con- 
stants. Deduce an expression for its acceleration at any time t. 
Also let c = 2, k = 3 , x and t being in feet and seconds respect- 
ively, and compute the acceleration when £ = 4 sees. 

10. The ''acceleration due to gravity" is about 32.2 ft.-per- 
sec.-per-sec. Express the same in yard-minute units. 

11. The law of a motion is a=io/ (ft.-sec. units), and the 
velocity is 10 ft.-per-sec. when tis 4 sees. Determine the veloc- 
ity at any instant. 

Solution : Since a = dv/dt =iot, dv = iot dt, 

and v = 1 o / 1 dt = $t 2 + C. 

Now v = 10 and / = 4 being simultaneous values of v and t, they 
satisfy the last equation; hence 

io = $X4 2 + C\ or C=—'jo } 
and v = $i 2 —yo. 

12. If the law of a motion is a = 10/ + 5 (ft.-sec. units) and x, 
v, and t are simultaneously zero, determine the values of v and x 
at any time. 

176. "Acceleration-Time" and Other Curves. — If simulta- 
neous values of- the acceleration and the time of a motion be 
plotted on two rectangular axes, then all such plotted points 
determine a curve called the acceleration-time curve for the 
motion. Evidently it is a graphical representation of the way 
in which the acceleration varies with the time. 

Other curves descriptive of a motion can be drawn. Thus 
values of the velocity (v) and the position-abscissa (x) of a 
moving point if plotted make a " velocity-space curve," and 
values of a and x if plotted make an "acceleration-space 
curve." 



176 RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 



§ II. Important Special Motions. 

177. Uniform Motion. — The velocity is constant, and there- 
fore the acceleration is zero and the displacement (x 2 — *i) i& 
any interval (t 2 — t t ) given by 

(x 2 -x 1 )=v(l 2 -t 1 ) > 

v denoting the velocity. 

178. Uniformly Accelerated Motion. — The acceleration is 

constant, and the velocity increment {v 2 — v^) in any interval 
(t 2 -t x ) is given by 

(v 2 -v 1 )=a(t 2 -t 1 ), (1) 

a denoting the acceleration. According to this equation the 
velocity varies uniformly; hence the average velocity for the 
interval (t 2 — t x ) is i(v 2 +v 1 ) t and the displacement (x 2 — x 1 ) in the 
interval is given by 

(x 2 -x 1 ) = ^(v 2 + v 1 )(t 2 -t 1 ) (2) 

Let x , v , and be simultaneous values of x, v, and t. Since 

a = dv/dt, dv = adt, or v = at + C v 

Substituting simultaneous values of v and / in the last equation, 
we find that C 1 = v ; hence 

v = at + v (3) 

Since v = dx/dt } dx = at dt+v dt; and hence 

x = iat 2 + v t + C 2 . 

Substituting simultaneous values of x and t in the last equation, 
we find that C 2 =x ; hence 

x = %at 2 + v t + x (4) 

EXAMPLES. 

A body moving near the earth under the influence of its 
attraction would have a constant acceleration were it not for 
air resistance. In the following examples neglect this resist- 
ance and denote the acceleration by g and the distance (meas- 



§11.] IMPORTANT SPECIAL MOTIONS. 177 

ured positively downward) of the moving body from the start- 
ing-point by x. 

i. If a body falls from rest, show that 

v = gt, x = ^gt 2 , and v 2 = 2gx. 

2. If a body is projected down with a velocity v^, show that 

v=gt+v , x = hgt 2 + v t, and v 2 = 2gx + v 2 . 

3. If a body is projected up with a velocity v 0l show that 

v=gt—v Q , oc = ^gt 2 —v Q t 1 and v 2 = 2gx + v 2 . 

179. Simple Harmonic Motion. — This may be defined as a 
rectilinear motion, in which the acceleration of the moving point 
is proportional to its distance from an origin in the path and is 
always directed from the point to the origin. It may also be 
defined thus: If a point travels in a circle describing equal dis- 
tances in all equal intervals of time, then the motion of the pro- 
jection of the point on any diameter of the circle is a simple har- 
monic one. We will choose the latter definition and show in the 
sequel that it is in accord with the former. 

Imagine P (fig. 152a) to start from P and move uniformly 
in the circle in the counter-clockwise direction. The motion of 
the projection of P on O x Y is harmonic and will now be discussed. 

Let e denote the angle XO^Pq, the lead (lag if negative) ; 
(0+e) " " " XO x P, the phase; 

y " ordinate O x V, the displacement; 

T ' time of one revolution of P, the period; 

n " " number of revolutions per unit time, the fre- 
quency; 

r " " radius of the circle, the amplitude; 

v " velocity of the projection (V) of P\ 

t " " time elapsed after starting. 
If d is expressed in radians, it is plain that 

27T 
d = —t = 2nnt = Ojt ) 

being an abbreviation for 2iz/T and 2-n; w may be defined 



i 7 8 



RECTILINEAR MOTION OF A PARTICLE. [Chap. VII- 



also as the angle described by O x P per unit time. From the fig- 
ure it will be seen that 



y = r sin (cot + e). 



and since v = dy/dt t 



v = cor cos (cot -\- s) = cor sin (cot -{- e + n / 2) 
= cox = co(r 2 — y 2 )% . 

Since a = dv/dt, 



a= —co 2 r sin (cot -he) = co 2 r sin (cot + s + ~) 



co 2 y. 



(a) 





4 


V 


Y 


-\ 2 

> \P 
-dC—^r 




6 / 








X if 


1 X 


8 \ 




°i 




H 


/ U 




10^ 






-"12 





(1) 
(2) 

(3) 




1 (b) 



if to 



1 (f) 



Equation (1) is represented in fig. 152(6). The curve is a 
sinusoid, and since the ordinates and abscissas denote "dis- 
placement" and time respectively, it is a displacement-time, 
or a space-time curve. The curve may be constructed as fol- 
lows: Having drawn the "circle of reference" (fig. 152a) and 
having fixed the initial position of P, divide the circumference 
into a number of equal parts (16 is convenient) beginning with 
P , and number that point o and the others successively in the 



I II.] IMPORTANT SPECIAL MOTIONS. i?9 

direction of the motion of P, i, 2, etc., 16 coinciding with o. 
Then on the t axis (fig. 1526) lay off any convenient length to 
represent the period, divide it in 16 equal parts, and erect ordi- 
nates at the points of division. Number these beginning with 
■O'y, o, 1, 2, 3, etc., the last being 16. Then project points o, 
1,2, etc., of the circumference of the circle upon the correspond- 
ing ordinates. The curve through the projections is the dis- 
placement-time curve. 

Equations (2) are represented in fig. 152 (c) and (d), (c) rep- 
resenting v = cox and (d) the other one. The equation v = a>% 
shows that the velocity at any displacement y equals co times 
the corresponding value of x. Hence to construct the curve in 
(c) lay off 2 V equal to O x V in (a) and make Vv equal to coO^H, 
laying it off up or down according as X H (x) is positive or nega- 
tive. Repeat this construction for several positions of P and 
thus determine the curve. 

The curve in (d) is the velocity-time curve for the harmonic 
motion and may be constructed in various ways: for instance, 
lay off 0"i6 to represent the period, divide it into sixteen equal 
parts, and number the points of division beginning with 0", o, 
1,2. etc. At these points erect values of the velocity which may 
be obtained from (c) by obvious methods. 

It is worth noting that the equation v = cor sin (cot + s + n / '2) 
is very similar to y = r sin (cut+s), v, cor, and s+tt/2 in the first 
replacing y, r, and = in the second. Hence the variation in v 
is simply harmonic and can be represented by a harmonic motion 
whose period, phase, and amplitude are respectively equal to, 
90 ahead of, and oj times the period, phase, and amplitude of 
the given motion. 

Since (b) and (d) are y-t and v-t curves respectively and 
v = dy/dt, any ordinate in the latter figure should be equal (by 
scale) to the slope of the tangent to the curve in the former at 
the corresponding point. Thus the ordinate Vv in (d) should 
equal the slope of the tangent at V in (6). 

Equations (j) are represented in fig. 152 (e) and (/), (<?) rep- 
resenting a= —co 2 y and (/) the other one. Equation a= —co 2 y 
shows that the acceleration is proportional to the displacement 
:(y) and that it is always opposite to the displacement in sign, i.e., 



180 RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 

it is always directed from the moving point toward the. middle 
of the path. Hence the two definitions of simple harmonic 
motion previously given agree. 

To construct the curve in (e), lay off 3 V equal to y and make 
Va equal to coy, laying it off down or up according as y is posi- 
tive or negative. Repeat this construction for a number of 
positions of P and thus determine the line s a. 

The curve in (/) is the acceleration-time curve for the har- 
monic motion and may be constructed in various ways: For 
instance, lay off 0"'i6 to represent the period, divide it into 16 
equal parts, and number the points of division, beginning with 
0"' , o, i, 2, etc. At these points draw ordinates to represent 
the corresponding values of the acceleration which may be ob- 
tained from (e) by obvious methods. 

The equation a = ^» 2 rsin (cot-\-£ + 7r) is the same in form as 
y = r sin (cot + e), a, oj 2 r, and (s+n) in the first replacing y, r, 
and s in the second. Hence the variation of a is analogous to 
that of y; in fact the variation in a is simply harmonic and 
it can be represented by a harmonic motion whose period, 
phase, and amplitude are respectively equal to, t8o° ahead of, 
and co times the period, phase, and amplitude of the given motion. 
Since (d) and (/) are v-t and a-t curves respectively and 
a = dv/dt, any ordinate in the latter figure equals (by scale) the 
slope of the tangent to the curve in the former at the corre- 
sponding point. Thus the ordinate Va in (/) equals the slope 
of the tangent at v in (d). 

1 80. Mechanism for Producing a Simple 
Harmonic Motion. — The mechanism repre- 
sented in fig. 153 consists of a crank and a 
slotted slider, the pin of the crank fitting 
the slot of the slider. As the crank is 
rotated the slider moves up or down, and 
plainly if the crank rotates uniformly every 
point of the slider executes a simple har- 
monic motion. 

The space-time curve for the motion of 
the slotted slider can be automatically 
drawn as follows : Fasten a pencil to the slider so that it will 



§n.] 



IMPORTANT SPECIAL MOTIONS, 



181 



mark on a plane surface as the slider moves, and then cause a 
sheet of paper to move uniformly over the surface in a direction 
at right angles to that of the motion of the pencil; the line 
traced on the paper is the space-time curve. If the mechanism 
for moving the paper is connected with the crank-shaft, then 
the curve traced by the pencil is a space-time curve for a har- 
monic motion whether the mechanism is driven uniformly or 
not. 

181. Motion of the Piston of a Steam-engine. — Let OP (fig. 
154) represent the crank and the connecting-rod of a steam T 
engine slider-crank mechanism, and suppose 
that the crank turns uniformly. 

Let n denote number of revolutions of the 
crank per unit time ; 
c " • length of crank; 
r " " " connecting-rod; 

y " the distance of C from ; 
t " " time required to describe 
the angle P OP. 
Then d = 2-ut = cot, to being an abbreviation 
for 27W, and Fig. 154. 

y = (r 2 — c 2 cos 2 cot)* + c sin cot (1) 

Hence the velocity of the piston is 

sin 2 cot 




dy 

j- =cco[ cos cot-\- 



and its acceleration is 

= — ceo 2 1 si 



d 2 y 
cW 



sin a>t + 



2 (r 2 /c 2 — cos 2 tot)*) 

cos 4 cot — (r 2 /c 2 ) cos 2 coi 
(r 2 /c 2 — cos 2 tot)* 



)■ 



(2) 



(3) 



Close approximate values of the velocity and acceleration can 
be found as follows : Since approximately 



|i~cos 2 ^j =r^i-^- 2 cos 2 ^j, 



*-('-£) 



H sin (2cot + 7i / 2 ) + c sin cot, very nearly. (4) 



182 RECTILINEAR MOTION OF A PARTICLE. [Chap. VII. 

Hence, approximately, the velocity of the piston is 

dy 

— = coj[cos wt + (c/2r) cos (2^ + 71/2)], . . . (5) 

and the acceleration is 
d 2 y 



dt'< 



= —ceo 2 [sin cut — (c/r) sin (2 wt + x/ 2)]. . . (6) 



If the connecting-rod were infinitely long, c/r would be zero 
and the second term of eq. (6) would vanish; hence the motion 
of the piston would be simply harmonic. The smaller the ratio 
c/r the more nearly is the motion of an actual mechanism sim- 
ply harmonic. 

EXAMPLE. 

Take r/c equal to four and compute the values of the accel- 
eration in terms of coo 2 from equations (3), (6), and (6) with the 
last term omitted, when = o°, ±30°, ±6o°, and ±90° (see fig. 
154). Plot these values on the same base, draw the three a-d 
curves, and compare. 



CHAPTER VIII. 
CURVILINEAR MOTION. 

§ I. Velocity and Acceleration. 

182. Specification of Position. — It is usually convenient here- 
in to specify the position of a point in space by Cartesian coor- 
dinates, but for the purpose of this section it is more convenient 
to specify position by means of a vector. The vector drawn 
from a fixed origin of reference to the point to be located is called 
the position-vector of the point. For, if the vector is known the 
position of the point with reference 
to the origin is also known. Thus, 
the direction of the vector OP (fig. 
155) fixes the direction of P from the 
origin 0, the length of the vector fixes 
the distance of P from O, and thus the 
position of P is determined. 

The position of a point in a given Fig. 1 55- 

line can be specified by a single quantity : Thus if A is an origin 
of reference in the line AP, the abscissa 5 measured from A along 
the line fixes P, it being understood that 5 is positive for points 
on one side of A and negative for those on the other. 

183. Space-Time Curve. — A curvilinear motion can be rep- 
resented in part by means of a line, the space-time curve for the 
motion. The ordinate and abscissa of each point of it repre- 
sent respectively simultaneous values of the position abscissa s 
of the moving point and of the time t. It is analogous to the 
space-time curve for a rectilinear motion (art. 164) and is simi- 
larly constructed. 

This curve must not be confounded with the path; the latter 
may be tortuous, but the former is always a plane curve. 

184. Displacement. — The displacement of a point during an 
interval in which it moves from A to B is defined to be the vec- 

183 



1 84 CURVILINEAR MOTION. [Chap. VIII. 

tor AB. The term therefore does not refer to the path actually 
described between A and B, but simply to the straight path 
between A and B. Observe that if is any point of reference, 
vectorially 

AB = OB-OA. 

Two displacements must agree in direction as well as in magni- 
tude in order to be equal. 

185. Kinds of Motion. — The definitions under this title given 
in art. 166 refer to rectilinear motion ; those following are general. 

If the displacements of a moving point in equal intervals of 
time (large or small) are equal, its motion is called uniform; and 
if unequal, the motion is called non-uniform. 

In order that all displacements may be equal, the path must 
be straight; hence a uniform motion as here defined must be 
rectilinear. 

186. Velocity. — By velocity of a point is meant the rate with 
respect to time at which it changes position or at which its dis- 
placement occurs. Still otherwise stated, velocity is the time- 
rate (of change) of its position- vector. 

Let -P(fig. 1 56) be a point moving in the path APB, and O 

and O' points of reference ; then OP is the 

,f/ // \ position- vector of P, and s its position-ab- 

,p^ \ scissa. Now in general the rate of the vec- 

/ \ \ tor OP, i.e., the velocity, changes with the 

I \ \ time as shown in Appendix B, and when the 

j \ \ moving point is at P the rate is a vector 

\\ ■ \ \ whose direction is that of the tangent at P, 

^° whose magnitude equals the time-rate of 
Fig. 156. (change of) 5. 

If the magnitude of the velocity* be denoted by v, then 

v = ds/dt; (1) 

and if s varies uniformly, ds/dt is constant and 

v = Js/Jt (2) 

* The word "speed" has been set apart by several recent writers to 
denote magnitude of a velocity. Such usage is very convenient and will 
be followed herein. 



§!•] 



VELOCITY AND ACCELERATION. 



185 



Observe that our definitions lead to the result that the veloc- 
ity in a curvilinear motion cannot be constant, for although its 
magnitude might be constant its direction continually changes. 
This result is in accord with the definitions of uniform and non- 
uniform motion (art. 185). 




EXAMPLE. 

The point P (fig. 157) describes the circle in such a way that 
5 = 2t 3 , s and t being in ft. and sees, respect- 
ively, (a) Deduce an expression for the 
speed at any time t. (6) What are the 
magnitude and direction of the velocity 
when t = 2 sees. ? 

Solution: (a) Since the speed (v) 
equals ds/dt, from the equation of the 
motion, v = 6t 2 . Fig. 157. 

187. Speed-Time Curve and Hodograph. — The way in which 
the speed of a moving point varies can be represented by a line 
called the speed-time curve for the motion. The ordinate and 
abscissa of each point of it represent simultaneous values of the 
speed and the time respectively. It is analogous to the veloc- 
ity-time curve for a rectilinear motion (art. 170) and is similarly 
constructed. 

If from any point vectors be drawn which represent the 
velocities of a moving point and the free ends of all such vectors 






be joined, the joining line is called the hodograph for the motion. 
Thus let P lf P 4 (fig. 158) be the path of a moving point, and let. 
its velocities at the points P lt P 2 , etc., be as represented by 



1 86 CURVILINEAR MOTION. [Chap. VIII. 

P x v x , P 2 v 2 , etc. The hodograph is P/P/, O'P/ being equal and 
parallel to P^, 0'P 2 f parallel and equal to P 2 v 2 , etc. 

A hodograph may be plane or tortuous, but a speed-time 
curve is always plane. The latter curve corresponding to fig. 
158 (a) and (b) is shown at (c). 

EXAMPLES. 

1. Suppose that P (fig. 157) describes the circle with a con- 
stant speed of 10 ft.-per-sec. Draw a hodograph and a speed- 
time curve for the motion from A to B. 

2. Draw a hodograph and a speed-time curve for a motion 
from A to B (fig. 157) according to the law stated in the ex. of 
art. 186, determining at least four points on the hodograph. 

188. Acceleration. — By acceleration of a point is meant the 
time-rate (of change) of its velocity. 

In deducing the expression for this rate it must be remem- 
, bered that velocity is a vector 

quantity. According to Appen- 
dix B the rate is a vector, and if 
A f P' (fig. 159) is a hodograph 
and O'P' represents the velocity 
a at any instant t, then the accel- 
FlG - 1 59- eration at that instant is a vector 

whose direction is that of the tangent to the hodograph at P' , 
whose magnitude is the time-rate of 5', 
s' being the distance of P' from any fixed point on the hodo- 
graph as origin.* 

If a be used to denote the magnitude of the acceleration, 

a = ds'/dt; (1) 

and if a varies uniformly, 

a=Js'/Jt (2) 

Observe carefully that the tangents to the hodograph and to 
the path at corresponding points P and P' are not parallel or in 

* As explained in the appendix referred to, the fixed point A' from 
which s' is measured is to be taken so that P' moves in the positive s' 
direction and the arrow on the tangent (giving the sense of the vector) 
points in the same direction. 





§1.] 



VELOCITY AND ACCELERATION. 



187 



any way related as to direction, but the vector representing the 
acceleration always points from P toward the side of the tan- 
gent on which the path lies. 

EXAMPLE. 

1. What is the acceleration of a point describing a circle of 
radius r with constant speed v} 

Solution: Let AB (fig. 160) be the circular path and P the 
position of the moving point at any instant. When P is at A 




A B 



Fig. 160. 




the velocity is represented by the vector O'A', when at P by O'P', 
when at B by O'B' , etc. Hence the hodograph is the circular 
arc A'P'B', its radius being v. Now the angles POA and P'O'A' 
are equal, and since the former equals s/r and the latter s'/v, 



Hence 



' s/r = s'/v t or s'=sv/r. 

ds'/dt = v/r{ds/dt) = v 2 /r. Since ds'/dt = a, 



= 7.2 



v 2 /r. 



The direction of the acceleration of P is that of the tangent 
to the hodograph at P'. Since this tangent is parallel to the 
normal to the path at P, 

the acceleration of a point describing a circle with 
constant speed is directed along the radius of the 
circle drawn to the moving point, and its value 
equals the speed squared divided by the length of 
the radius. 



1 88 CURVILINEAR MOTION, [Chap. VIII. 



§ II. Resolutions of Velocities and Accelerations. 

189. Components and Resultant of Velocities or Accelerations 
Defined. — Velocities and accelerations are, as defined in arts. 
186 and 188, vector quantities; they may therefore be resolved 
and compounded. The velocity (or acceleration) represented 
by the sum Of the vectors representing any number of velocities 
(or accelerations) is called the resultant of those velocities (or 
accelerations). The velocities (or accelerations) represented by 
the components of a vector representing any velocity (or accel- 
eration) are called the components of that velocity (or accelera- 
tion). 

190. Axial Components of a Velocity. — These components are 
parallel to three rectangular axes(#, y, and z), and will be de- 
noted by v x , v y , and v z respectively. 

Let v denote a velocity, its direction angles with the x, y, 
and z axes being a, /?, and y respectively. Then the x, y, 
and z components equal respectively v cos a, v cos /?, and v cos 7-. 
Since v = ds/dt, cos a = dx/ds, cos ft = dy/ds, and cos y = dz/ds, 

v x = dx/dt, v 2/ = dy/dt, v z = dz/dt. . . . (1) 

From these equations it is plain that the x, y, and z components 
of a moving point are respectively the time-rates at which its 
distances from the yz, zx, and xy planes change, and that they 
equal respectively the velocities of the projections of the moving 
point on the axes x, y, and z. 

191. Tangential and Normal Components of Velocity. — These 
components are parallel to the tangent and normal to the path 
at the point where the velocity is resolved. They will be de- 
noted by v t and v n respectively. Since the velocity at any point 
in the path is directed along the tangent to the path at that 
point, 

v,=v and v„ = o (1) 

192. Axial Components of Acceleration. — These components 
are parallel to three coordinate axes (x, y, and z), and will be de- 
noted by a x , a y , and a z respectively. For simplicity, the deduc- 
tion of the expressions for the components is limited to the case 



§ II.] RESOLUTIONS OF VELOCITIES AND ACCELERATIONS. 189 

of plane paths with the coordinate axes in the plane of the path, 
but the discussion can be extended to the general case (resolu- 
tion of the acceleration of a point along three axes, the path 
being tortuous). 

Let PQ (fig. 161) be the path of a moving point and P'O' the 
hodograph of the motion, P and P' being any two corresponding 




(a) ( b ) 

Fig. 161. 

points on the path and hodograph respectively. Also let v and 
a denote the magnitude of the velocity at P and its angle with 
the x axis respectively. Then the polar coordinates of P' are 
v and a, and its rectangular coordinates are v cos a and v sin a, 
or v x and v y ; hence the coordinate axes in fig. 161(6) are marked 
0'V X and 0'V y instead of OX and OY. 

As shown in art. 188, the acceleration at P is directed along 
the tangent to the hodograph at P f ; hence P'a may represent 
the acceleration. Now if a, a x , and a y denote the acceleration 
and its x and y components respectively, it follows from the 
figure that 



a cos a 



and 



a sin a 



and from the analogy between the two parts of the figure, 
dv x /ds' and sin a' = dv y /ds' , 



cos a 



ds' being the length of the infinitesimal arc on the hodograph at 
P f . Also, from art. 188, a = ds'/dt; hence 

a x = (ds'/dt) (dv x /ds') = dv x /dt, 

and a y = (ds f /dt)(dv y /ds') = dv y /dt. 

In the general case (resolution into three components) it can be 



190 CURVILINEAR MOTION. [Chap. VIII. 

shown that the z component of the acceleration (a z ) equals 
chg/dt; hence 

a x = dv x /dt = d 2 x/dt 2 ;) 

a,, = dv y /dt = d 2 y/dt a ; V (i) 

a, = dv,/dt = d 2 z/dt 2 . ) 

The equations state that the x, y, and z components of the 
acceleration of a moving point P equal the time-rates of the 
x, y, and z components of its velocity; also that they equal 
the accelerations of the projections of the point on the x, y, 
and z axes respectively. 

193. Tangential and Normal Components of Acceleration. — 
These are components whose directions are parallel to the 
tangent and normal to the path at the point where the accel- 
eration is resolved; they will be denoted by a t and a n re- 
spectively. For simplicity, the deduction below is limited 

to the case of a plane path, but it 
might be extended to the general 
case. Let a denote the acceleration 
of a moving point when it arrives at 
P (fig. 162). Since the values of the 
tangential and normal components 
F IG - l62 - of the acceleration are independent 

of the axes of reference,* they may be chosen parallel to the 
tangent and normal at P. This is done solely for simplicity in 
the deduction of the values of a t and a n . With axes thus chosen, 
it is plain that the tangential and x components and the normal 
and y components are equal, or 

a t = a x = dv x /dt, a n = a t = d 2 y/dt 2 . 

Now d 2 y/dt 2 = (d 2 y/dx 2 )/(dx 2 /dt 2 ) = (d 2 y/dx 2 )v x 2 ; also, cPy/dx 2 
equals the curvature of the path at P. Denoting the radius of 
curvature at P by p, and since v x = v t = v, 

a, = dv/dt = d 2 s/dt 2 , and a n = v 2 />. . . . (1) 

* This independence may be explained thus: From fig. 161 it is plain 
that fl ( =acos {a — a') and a n =a sin (a — a'). Now both a and (a — a') 
are independent of the coordinate axes and hence a t and a n are also. 




§ II. J RESOLUTIONS OF VELOCITIES AND ACCELERATIONS. 191 

These equations state that the tangential and normal accelera- 
tions at P (any point in the path) equal respectively the time- 
rate (of change) of the speed at P and the square of the speed 
divided by the radius of curvature at P. 

If the moving point travels in a circle of radius r and with 
a constant speed v, then dv/dt = o and 

a = a n = v 2 /r (2) 

EXAMPLES. 

i. A point P (fig. 163a) starts from a point X and moves in 
a circle of 20 ft. radius and so that the distance described (in 
feet) equals twice the cube of the time (in seconds) after start- 





ed) 

Fig. 163. 

ing. Compute the value of v, v u v n , v x , v y , a, a t , a n , a x , and a y , 
when t = 2 sees. 

Solution: From the equation of motion, s = 2t s - y 

.-. ds/dt = 6t 2 = v, d 2 s/dt 2 = i2t = a t ; also, a n = v 2 /2o. 

Hence, when t = 2 sees., 

£' = 24 ft. /sec., v =24 ft./sec, v n = o, 

a, = 24 ft. /sec. 2 , and a n = 28.8 ft./sec. 2 . 

Since a: = 20 cos 6 and ^=20 sin d, 

dx/dt = - 20 sin 6 dd/dt = v x , dy'/dt = 20 cos 6 dd/dt = v y , 
d 2 x/dt 2 = -20 sin d 2 0/dt 2 -2o(dd/dt) 2 cos d = a x , 
and 

d 2 y/dt 2 = 20 cos d 2 0/dt 2 -2o(dd/dt) 2 sin d = a y . 

Since 6 = s/2o = o.it B , dd/dt = o.^t 2 , and d 2 6/dt 2 = o.6t. By means 
of these values and the expressions for v x ,v y , a x , and a y , we find 
that when t = 2 sees., 



I9 2 CURVILINEAR MOTION. [Chap. VIII„ 

v x = — 17.21 and v y = 16.7 ft. /sec, 
a x = —37.2 and a^ = —3.9 ft. /sec. 2 . 

The determination of the magnitude and direction of a is left to» 
the student; he should also plot to scale, in the figure, all the 
computed values. 

2. Discuss the motion of the centre of gravity of a projectile 
neglecting the resistance of the air if the velocity and angle of 
projection are v' and e respectively (see fig. 1636). 

Solution: It is shown in art. 240 that the acceleration of the 
centre of gravity at each instant is just like that of a freely fall- 
ing body, i.e., equal to g and vertically downward; hence 

a x = o and a y =-g. (1) 

Since d 2 x/dt 2 = o,v x = C 1 and x = C x t + C 2 , C\ and C 2 being con- 
stants of integration. Since v x is constant during the entire 
motion, it is equal to its initial value, i.e., C 1 = v f cos e, and since 
x = o when t = o, C 2 = o ; hence 

v = n f cos e and x = v' cos e-t (2) 

Since d 2 y/dt 2 = — g, v y = — gt + C 3 and y = — \gt 2 + C 3 t + C ir 
C 3 and C 4 being constants of integration determinable like C t 
and C 2 from ' ' initial conditions . ' ' These are y — o and v s =v' sin e 
when t = o ; hence C 3 = v' sin s and C 4 = o, and therefore 

v y = — gt + v f sin e and y= —igt 2 + v' sin s-t. . (3) 

3 . Show that the path is parabolic and that its equation is 

j = x tan £ — gx 2 1 2V 2 cos s. 

4. Deduce expressions for the range and greatest height 
reached, and show what values of e make them a maximum. 

§ III. Relativity of Motion. 

194. Meaning of Relative Path, Displacement, Velocity, and 
Acceleration. — The position of a point can be specified only with 
respect to a set of reference axes or some other equivalent base, 
i.e., specification of position is necessarily relative. This fact 
makes the path, displacement, velocity, and acceleration of a 
moving point relative, as can be clearly shown by illustration. 



§ III.] RELATIVITY OF MOTION. 193 

Imagine a drawing-board and a sheet of paper lying upon it. 
Imagine also that a bug walks about on the paper, that the latter 
is slid about on the board, and that the bug can and does punch 
holes rapidly through the paper into the board. The succession 
of holes in the paper marks out the path of the bug relative to 
any axes on the paper, and the holes in the board mark out its 
path relative to axes on the board. In general these two paths 
would be very dissimilar. 

Again, imagine a running locomotive and consider the motion 
of a point on the rim of one of its wheels. Relative to a set of 
axes on the wheel, the position of the point is fixed, i.e., the 
point does not move ; relative to axes on the body of the loco- 
motive, the point describes a circle; and relative to axes on the 
earth, the point describes cycloids. 

It is evident now that displacement is relative and that the 
displacements of a moving point relative to different axes are in 
general different. Thus referring to the first illustration above, 
let A and A' be holes punched in the paper and board at the 
same instant, and B and B' holes punched at a later instant. 
Then the displacements relative to the paper and board are rep- 
resented by the vectors AB and A'B' respectively. 

According to the preceding and definitions (arts. 186 and 188) , 
velocity and acceleration are also relative. A velocity com- 
puted from displacements relative to a certain set of axes must 
be regarded as relative to that set, and an acceleration com- 
puted from velocities relative to a certain set of axes must be 
regarded as relative to that set. Like displacements, the veloc- 
ities and accelerations of a moving point relative to different 
sets of axes are in general unlike. 

195. Definitions. — By path, displacement, velocity, and ac- 
celeration of a point with respect to a rigid body is meant its 
path, displacement, velocity, and acceleration respectively with 
respect to any set of axes fixed in the body. This is definite, 
for the paths, displacements, velocities, or accelerations of a 
point relative to different sets of axes in the same rigid body are 
the same. 

By path, displacement, velocity, and acceleration of a moving 
point with respect to another point is meant its path, displace- 



194 



CURVILINEAR MOTION. 



[Chap. VIII. 



ment, velocity, or acceleration relative to axes of fixed directions 
passing through the second point. This is also definite, for the 
path, displacements, velocities, or accelerations of a point with 
respect to different sets of axes having the same origin and fixed 
directions are the same. 

When the motion of a point is referred to two sets of axes, 
two points, or two bodies, it is sometimes convenient to call one 
of the motions " absolute" and the other " relative." 

196. Relation between the Velocities and Accelerations of a 
Point Relative to Two Points. — Proposition. — The velocity (or 
acceleration) of A relative to B equals the vector sum of the 
velocity (or acceleration) of A relative to C and that of C rela- 
tive to B. 

Proof (restricted to the case in which the three points A, B, 
and C are continually in the same plane) : Employing the first 
illustration of art. 194, let the punching bug be A, and the lower 
left-hand corners of the board and paper be B and C respectively. 
Also let the edges of the board and paper through B and C be 
used as two sets of coordinate axes. In the motion the edges 
of the board and paper are to remain fixed in direction. Thus 
let UBV (fig. 164) represent the board and X 1 C 1 Y 1 and A t any 




Fig. 164. 

initial position of the paper and the bug, and X 2 C 2 Y 2 and A. 
their positions at the end of any interval of time. 



§ III.] RELATIVITY OF MOTION. 195 

The first hole punched in the paper is at A % * at the end of the 
interval, C 2 A X ' being equal and parallel to C^A^\ hence, 

relative to C, the displacement of A is A 1 , A 2 \ 

" B, " " " A is A X A 2 \ and 

11 B, " " . " CisC t C 2 . 

It is obvious from the figure that vectorially 

A 1 A 2 = A 1 A 2 -\-C 1 C 2 , 

i.e., the displacement of A relative to B equals the vector sum, 
of the displacement of A relative to C and that of C relative to 
B. It follows that the displacements per unit time are simi- 
larly related; and since displacement per unit time is velocity, 
the proposition as to velocity is proved. 

Let ac' and c'b' (fig. 165) represent the velocity of A relative 
to C and that of C relative to B at a given instant. According 
to the proof above, ab' represents the ve- 
locity of A relative to B at that instant. 
Let ac" and c"b" represent the velocity of 
A relative to C and that of C relative to B 
at a later instant . Then ab" represents the 
velocity of A relative to B at that instant. 
Now c'c" and b'b" represent respectively 
the increments in the velocity of A rela- 
tive to C and in the velocity of A rela- 
tive to B during the interval. If b'oc"c f 
is a parallelogram, then ob" represents 

the increment in the velocity of C relative to B. From the 
triangle ob'b" it is plain that vectorially 

b'b"=b'o+ob", 

i.e., the increment in the velocity of A relative to B equals the 
vector sum of the increment in the velocity of A relative to C 
and that of the velocity of C relative to B. It follows that the 
increments per unit time are similarly related, and since the 
increments (or velocity changes) per unit time are accelerations, 
the proposition is proved. 

197. Meaning of Composition of Motions. — According to the 
preceding article the displacement of the bug (of the first illustra- 




196 



CURVILINEAR MOTION. 



[Chap. VIII. 



tion of art. .194) with respect to a point (B) on the board may 
be found by compounding its displacement with respect to a 
point (C) on the paper and that of C with respect to B. Hence 
we say that the motion of a point A with respect to a point B 
may be regarded as consisting of the motion of A with respect 
to C and that of C with respect to B. 

EXAMPLES. 

1 . A man rows a boat in a stream whose velocity is 3 miles 
per hour, so that his velocity relative to a floating chip is 5 mi.- 
per-hr. "straight across" the stream. Determine his absolute 
velocity.* 

2. Two locomotives run at speeds of 40 and 50 mi.-per-hr. on 
two different tracks, that of the first locomotive being east and 
west and the other northeast and southwest. If both locomotives 
run eastward, what is the velocity of each relative to the other? 



§ IV. Composition of Simple Harmonic Motions. 

198. Mechanism for Compounding Simple Harmonic Motions, 

— Fig. 166(a) represents two cranks with their slotted sliders, 
the crank-shaft of one turning in a bearing fixed upon the slider 
of the second. If both cranks be turned uniformly, the lower 




slider S' executes a s.h.m. relative to the support of the mechan- 
ism, the upper slider S" executes a s.h.m. relative to the lower 

* Motion referred to points on the earth is often called for convenience 
" absolute motion," and the corresponding velocities and acceleration are 
also called absolute. 



§ IV.] COMPOSITION OF SIMPLE HARMONIC MOTIONS. *97 

slider, and its motion relative to the fixed support is compounded 
of these two harmonic motions. In theory at least, a third crank 
and slider could be mounted on the second slider, a fourth crank 
and slider on the third slider, etc. Then if all cranks were turned 
uniformly, the motion of the last slider would be compounded 
of the simple harmonic motions executed by the several sliders 
relative to the supports of their cranks. 

When the sliders are in line the s.h.m.'s are described as 
"along the same line" or "collinear," and when the sliders are 
inclined to each other the s.h.m.'s may be described as "oblique " 
or " non-parallel." 

199. Composition of Two Collinear Simple Harmonic Motions 
of the Same Period.* — Let fig. 166 represent the position of the 
mechanism at any time t of the motion. We will consider the 
motion of the projection of C 2 on OY, i.e., V. 

From draw vectors OP' and OP" to represent the cranks 
as shown, and complete the parallelogram OP'P"P\ then OP is 
the sum of the vectors representing the cranks. It is plain from 
the figure that V is also the projection of P on the line OY. 
Since the periods of the two s.h.m.'s are the same, the angle 
P'OP" remains constant during the motion; hence OP is con- 
stant in length and turns uniformly, i.e., describes a circle at a 
uniform rate. Therefore the motion of V is simply harmonic. 

The amplitude of the motion of V is OP, the phase is XOP, 
the period is the same as that of the given s.h.m.'s. The epoch 
can be found from the figure as follows : Turn the parallelogram 
back to its position when t = o ; let P be the corresponding posi- 
tion of P, then XOP is the desired epoch. 

Values of the amplitude and epoch can be computed from the 
figure ; thus let 

a 1 denote the amplitude of the first s.h.m. 



a 2 


u << " 


" second " 


£1 


" epoch 


" first 


£2 


<< 


second " 


y* " 


" displacement 


' ' first 


y 2 " 


ii " >' 


second " 



* Only motions of the same period are compounded herein. 



I9 8 CURVILINEAR MOTION. [Chap. VIII. 

and 27: /(o their common period ; 

then y 1 = a x sin ( cot + sj and y 2 = a 2 sin ( cot + s 2 ) . 

Since the motion of V is simply harmonic, period equal to 
2n/co, its displacement is given by 

y = a sin (-orf+e);, 

a and £ being the amplitude and epoch respectively. It can be 
shown from the figure that 

a = a 1 2 + o 2 2 + 2a 1 o 2 cos (e 2 —£i), 

and tan $ = {a 1 sin £ t + a 2 sin £ 2 )/( a i cos £1 + a 2 cos £2) • 

200. Resolution of a Simple Harmonic Motion into Two Com- 
ponents Collinear with it. — Let OP (fig. 166) represent the crank 
of any s.h.m. in the line OY, and let OP' and OP" be the cranks 
of two others in the same line. According to the preceding ar- 
ticle the resultant of the second two gives the first and the latter 
are therefore called components of the first. 

Obviously a s.h.m. may be resolved into many pairs of com- 
ponents, for many parallelograms, as OP'P"P ', can be drawn on 
the same diagonal OP, and the sides OP' and OP" of each rep- 
resent the cranks of a pair of components. 

Special Case. — Resolution into two components which differ 

90 in phase, the epoch of one being zero. Let OP (fig. 167) 

represent the crank of the s.h.m. to be 

resolved when t = o; then XOP is the 

epoch of that motion and OP' and OP" 

x are the cranks of the component motions 

when t = o ; the epoch of one component 

Fig. 167. (crank OP') is zero and that of the other 

is 90 . Also if OP = a and XOP=--e, the equation of the given 

s.h.m. is 

y = a sin (ajt+e), 

and those of the first and second components are 

y' = a cos e • sin u)t, 

and y" = a sin s(sin cut + oo°) = a sin e • cos cot. 



IV.] COMPOSITION OF SIMPLE HARMONIC MOTIONS. 



199 



201. Composition of Many Collinear Simple Harmonic Motions 
of Equal Periods. — It follows from art. 199 that the resultant of 
three or more collinear s.h.m.'s of equal period is simply har- 
monic and of that period. For the first two can be replaced by a 
single s.h.m., and in turn this one and the third can be replaced 
by a single s.h.m., etc. 

To obtain the "crank" of the resultant motion, add the vec- 
tors' representing the several cranks in their positions at any 
instant ; the sum represents the crank of the resultant motion 
in its corresponding position. Thus, let OP', P'P", P"P'" , etc. 
(fig. 168), represent the cranks in their simultaneous positions; 




Fig. 168. 

then OP represents the length and direction of the crank of the 
resultant motion at the same instant. 

The epoch can be determined by turning the polygon about 
until P' falls into its position when t was zero. Let P be the 
corresponding position of P; then XOP is the epoch sought. 



EXAMPLES. 

1. Compound two collinear s.h.m.'s whose amplitudes and 
periods are equal, their phases differing (a) by 90 , (b) by 180 , 
(c) by 270 . 

2. Resolve the s.h.m. whose equation is y = 4 sin (27^ + 30°) 
into two collinear with it, their phase difference being 45 and 
the epoch of one being zero. 

3. Compound three collinear s.h.m.'s whose amplitudes and 
periods are equal, their phases differing by 120 . 



200 



CURVILINEAR MOTION. 



[Chap. VIII. 



H 



I x 



Q 



a # • 



i«-i 




/ '/ \ 



202. Composition of Two Simple Harmonic Motions in Lines 
at Right Angles. — Imagine the two sliders of fig. 166 turned at 

right angles as represented in fig. 169; 
then if the absolute motion of the 
lower one is simply harmonic and the 
motion of the upper one relative to 
the lower is also simply harmonic, the 
absolute motion of the upper is com- 
pounded of two s.h.m.'s at right 
angles. 

For simplicity, we choose the ori- 
gin of time so that the epoch of the 
first s.h.m. is zero and consider the 
motion of a point P at the middle of 
the slot of the second slider. With 
F IG - l6 9- notation and axes as in the figure, the 

x and y coordinates of P are 

x = a ± sin cot and y =- a 2 sin (cot + d). . . . (1) 

The equation of the path of P is found from these by eliminating 
i\ thus we find that 



v 






2 cos d , y 2 

xy + — 2 

a.a~ a 9 2 



sm' 



(»> 



This represents an ellipse and hence the motion has been 
called " elliptic harmonic motion." 

The path of a point whose motion is the resultant of two 
s.h.m.'s in lines at right angles (or inclined) can be traced 
graphically by plotting on the two lines simultaneous values of 
the displacements due to the component motions and then fix- 
ing the position of the point from the displacements. Thus let 
the circles of fig. 170 be the circles of reference of the two com- 
ponent s.h.m.'s, and let Y'Oa be the epoch of the horizontal 
and XOb that of the vertical component motion; then the dis- 
placements in the two motions when t = o are respectively Ox 
and Oy , and the position of the point describing the elliptic 
motion is c . The constructions for the other points on the 
ellipse should be obvious from the figure. 

Special Cases. — (1) If the phases of the component motions 



§ IV.] COMPOSITION OF SIMPLE HARMONIC MOTIONS. 



are the same or differ by 180 , sin d = o; hence according to 
eq. (2) the path is a straight line. The resultant motion is a 
s.h.m., its period equals that of the given motions, its phase is 




Fig. 170. 
the same as that of one or both of the given motions, and its 
amplitude equals (a x 2 + a 3 2 )*. Prove. 

(2) If the amplitudes of the component motions are equal 
and they differ in phase by 90 , a 1 = a 2 , cos^ = o, and sin d=i; 
hence according to eq. (2) the path is a circle whose radius 
equals a 1 = a 2 . The circular motion is "uniform," of a period 
equal to that of the component motions. Prove. 

203. Resolution of a S.H.M. into Two Components at Right 
Angles to Each Other. — Let OZ (fig. 171) be the path of the 
motion to be resolved, being the centre, and 
let the displacement, amplitude, period, and 
epoch be z, a, 2kJco, and e respectively; then 

z = a sin (tot+s). 

The x and y components of this displacement 
are (see the figure), 

x = (a cos a) sin (cot + e) 

and y = (a sin a) sin (cot + e); 

hence the periods and epochs of the components are the same 





x__ 


' \ 




zy 


1 
1 
1 
12/ 

1 
1 





/\ a 


1 

! 


/ 







202 



CURVILINEAR MOTION. 



[Chap. VIII- 



as of the given motion, and their amplitudes are respectively 
a cos a and a sin a. 

204. Composition of More than Two S.H.M.'s Not Collinear. — 

According to the preceding article each s.h.m. may be resolved and 
replaced by two components along two axes x and y. Accord- 
ing to art. 201 all the components in the x axis can be com- 
pounded into a single s.h.m. , and those in the y axis also. Accord- 
ing to art. 202 these two s.h.m. 's compound, in general, into an 
elliptic harmonic motion. 

Special Case. — Three s.h.m. 's in lines inclined 120 to each 
other of equal amplitudes and periods but differing 120 in 
phase, compound into a uniform circular motion. 

Proof : Let z'z" and z'" denote simultaneous displacements in 
the component motions and a and 271/ to their common amplitude 

and period ; then 

z' =a sin cot, 
z" = asin (cot -\-120 ), 
z'" = a sin (^ + 240°). 
Let OA, OB, and OC (fig. 172) be 
the paths of the^ component mo- 
tions, being the centre of each, 
x' , x" , and x'" the x components 
of the displacements, and y' , y" , 
and y"' their y components. Then, 
according to the preceding article, 




Fig. 172. 
x' = a sin (cot), 



y 



x" = acos i2o°-sin (^ + 120°), / / = asin 120°- sin O* + i20°), 
i"' = acos 240°- sin (cot + 2 40 ), /" = asin 240°- sin (a^ + 240 ); 
hence the sums of the x and y components (which call x and y) 
are 



x = &a sin cot, 



and -y = fa cos cot = % sin (cot + go°). 



These two equations show that the given motions are equivalent 
to two s.h.m. 's in the coordinate axes, equal as to amplitudes and 
periods but differing oo° in phase. According to special case (2 ) , 
art. 202, these two motions are equivalent to a uniform motion 
in a circle of radius fa, its period being equal to that of the 
given motions. 



CHAPTER IX. 
MOTION OF A RIGID BODY. 

§ I. Translation. 

205. Translation Defined. — A translation is a motion in which 
each straight line of the moving body remains fixed in direction. 
Thus the motions of the coupling-rods of a locomotive which runs 
on a straight track are translations ; also the motions of those 
on a locomotive which runs on transfer-table. 

206. Motions of all Points of a Body in Translation are Alike. 
— Let A and B be any two points of a body having a translatory 
motion, A' and B' their positions at a given instant, and A" and 
B" those at another instant. By definition, the lines A'B' and 
A"B" are parallel, and, since they are also equal in length, the 
figure A'B'A"B" is a parallelogram and A' A" and B'B" are 
equal and parallel. Hence the displacements of all points of 
the moving body for the same interval of time (long or short) 
are equal in magnitude and the same in direction. It follows 
that at each instant the velocities, and hence the accelerations, 
of all points of the moving body are exactly alike. 

207. Velocity and Acceleration of the Body. — By velocity 
and acceleration of a body having a translatory motion is meant 
the velocity and the acceleration respectively of any one of its 
points. 

§ II. Rotation. 

208. Rotation Defined. — A rotation is a motion in which one 
line of the moving body or of its extension remains fixed. The 
fixed line is called the axis of the rotation. 

Obviously all points of the moving body must describe cir- 
cles whose centres lie on the axis unless the axis cuts the body ; 
in this case all points of the body on that line are at rest, the 
others describing circles. The planes of the circles are perpen- 

203 



204 MOTION OF A RIGID BODY. [Chap. IX. 

dicular to the axis, and any plane perpendicular to the axis may 
be called the plane of rotation. All points of the body on any 
line parallel to the axis move alike, hence the motion of the pro- 
jection of the line on the plane of the motion represents that of 
all the points, and the motion of the body itself is represented by 
the motion of its projection. 

209. Angular Displacement. — By angular displacement of a 
rotating body during any time interval is meant the angle de- 
scribed during that interval by any line of the body perpendicular 
to the axis. Obviously all such lines describe equal angles in the 
same interval, and we select a line which cuts the axis. For con- 
venience, angular displacements are given sign — positive if dur- 
ing the interval the body has turned counter-clockwise, and 
negative if clockwise. 

Let the irregular outline (fig. 173) represent a rotating body, 
the plane of rotation being that of the 
paper, and the intersection of the 
axis with that plane. Let P be any 
point and 6 the angle XOP, OX being 
any fixed line of reference. As custom- 
arily, 6 is regarded as positive or nega- 
tive according as OX when turned 
about toward OP moves counter- 
IG ' I73 ' clockwise or clockwise. If 6 1 and 6 2 

denote initial and final values of 6 corresponding to any rota- 
tion, then the 

angular displacement = 6 2 — 6 1 = Ad. 

210. Angular Velocity. — The angular velocity of a rotating 
body is the time-rate at which its angular displacement occurs, 
or, otherwise stated, it is the time-rate at which any line of the 
body perpendicular to the axis describes angle. 

The time-rate at which OP (fig. 173) describes angle or the 
time-rate (of change) of 6 is, as shown in works on calculus and 
in Appendix B, dO/dt. Hence if w denotes angular velocity, 

co = d6/dt . (1) 

If the body turns uniformly, 6 is a uniform variable, and its time- 




§ II. J ROTATION. 205 

rate is Ad/ 'At, Ad denoting the angular displacement occurring in 
the interval At. Hence 

co = Ad/ At, (2) 

and the angular velocity is constant. 

211. Units of Angular Velocity. — The formulas of the pre- 
ceding article imply as unit an angular velocity corresponding 
to a unit angular displacement in each unit time, the velocity 
being constant. There are several such units ; thus, one revolu- 
tion-per-second, one degree-per-hour, one radian-per-second, etc. 
The last is the one usually used herein.* 

212. Sign of Angular Velocity. — An angular velocity must be 
regarded as having sign, the same as that of dQ/dt (and of Ad/ At 
if the angular velocity is constant). Now dd/dt and Ad /At are 
positive or negative according as d increases or decreases alge- 
braically ; hence 

The angular velocity of a rotating body at any in- 
stant is positive or negative according as it is 
turning in the counter-clockwise or clockwise 
direction at that time. 

213. Angular Acceleration. — The angular acceleration of a 
rotating body is the time-rate (of change) of its angular velocity. 

If, as in the preceding, co denotes the angular velocity, then 
the general expression for the time-rate of the angular velocity 
is dco/dt; hence if a denotes the angular acceleration, 

a = dco/dt = d 2 d/dt 2 (1) 

// the angular velocity changes uniformly, its time-rate is 
A co I At, A co denoting the increment in the velocity for any inter- 
val At; hence 

a = Aco/At, (2) 

and the angular acceleration is constant. 

2 14. Units of Angular Acceleration. — The formulas of the pre- 
ceding article imply as unit an angular acceleration correspond- 
ing to a unit angular velocity change in each unit time, the angu- 
lar acceleration being constant. One revolution-per-second- 

* For dimensions of a unit angular velocity, see Appendix C. 



206 MOTION OF A RIGID BODY. [Chap. IX. 

per-second, one radian-per-second-per-second, etc., are such 
units.* 

215. Sign of Angular Acceleration. — An angular acceleration 
must be regarded as having sign — the same as that of dco/dt 
(and of Aw/ At if the angular velocity changes uniformly). Now 
doj/dt and Aa>/At are positive or negative according as co in- 
creases or decreases algebraically; hence 

An angular acceleration is positive or negative ac- 
cording as the angular velocity is increasing or de- 
creasing (algebraically). 

216. Velocity and Acceleration of Any Point of a Rotating 
Body. — Let P (fig. 173) be any point of the rotating body there 
represented, let r denote its distance from the axis and 5 the 
length of the arc P P. Then if 6 is expressed in radians, s = rd; 
hence 

ds/dt = r dd/dt and d 2 s/dt 2 = r d 2 d/di 2 . 

Now ds/dt is the velocity of P (art. 186) and d 2 s/dt 2 is its tan- 
gential acceleration (art. 193); hence, if as heretofore the veloc- 
ity, the acceleration, and its tangential, and the normal com- 
ponents be denoted by v, a, a t , and a n respectively, 

y = rco, (1) 

a / = ra, a n = rw 2 , (2) 

and a = rVa 2 + w 4 ; (3) 

co and a denoting respectively the angular velocity and acceler- 
ation of the body. These equations show that the velocity 
and acceleration of a point in a rotating body are proportional 
to its distance from the axis. 

EXAMPLES. 

1. Write in the proper place below the signs of the angular 
velocity and acceleration of a body which rotates as follows: 

(a) Clockwise, 

(1) when "getting up speed," sign of co is . . . , of a . . . ; 

(2) " "slowing down," " """...,""...; 

* For dimensions of an angular acce/eration, see Appendix C. 



§ III.] PLANE MOTION. 207 

(6) Counter-clockwise, 

(1) when "getting up speed," sign of co is ..., of a ...: 

(2) " "slowing down," " """..., "".... 

2. Express an angular velocity of 11 rev.-per-sec. in rad.-per- 
sec. 

3. If the angular velocity of a wheel changes from 100 to 120 
rev.-per-min. in one-half a min., what is its average angular 
acceleration ? 

4. A wheel is set to rotating in such a manner that the num- 
ber of turns made after starting equals the square of the time (in 
minutes) after starting. Deduce expressions for the angular 
velocity and the acceleration at any time. 

Solution: The law of the motion, if d and t denote the num- 
ber of turns and the time, is d = t 2 ; hence 

dd/dt = 2t = a> (turns-per-min . ) , 

and d 2 6/dt 2 = 2 =a (turns-per-min. -per-min.). 

5. Compute the velocity and the acceleration of a point on 
the rim of a wheel whose diameter is three feet, if its angular 
velocity is 4 rev.-per-min. 

Solution: ^ = 4X6.283 = 25.13 rad./sec. 

According to eq. ( 1 ), v = ^co = 75.4 ft. /sec, and 

" eql(2), a n = v 2 /s = 1895 ft./sec 2 

Since the angular velocity is constant, the angular acceleration 
is zero; hence, see eqs. (2) and (3), a t = o and a = a n . The direc- 
tion of a is from the point on the rim to the centre of the wheel. 

§ III. Plane Motion. 

217. Plane Motion Denned. — A plane motion is one in which 
each point of the moving body remains at a constant distance 
from a fixed plane. The fixed plane (or any plane parallel to it) 
is called the plane of the motion. 

The wheel of a car running on a straight track has plane 
motion; so also has a book sliding about on a table. A trans- 
lation may or may not be a plane motion (see illustrations, art. 
205), but a rotation is always a plane motion. 

As in a rotation, all points on any line of the moving body 



2o8 MOTION OF A RIGID BODY. [Chap. IX. 

which is perpendicular to the plane of the motion move alike, 
hence the motion of the projection of the line on the plane of 
the motion represents that of all the points, and the motion of 
the body itself is completely represented by that of its projec- 
tion on the plane of the motion. 

218. Angular Displacement. — By angular displacement of a 
body whose motion is plane is meant (as in rotations) the angle 
described by any line of the body which is perpendicular to the 
plane of the motion. Obviously all such lines describe equal 
angles in the same time interval. As in rotations also, displace- 
ments are regarded as positive or negative according as they 
are due to counter-clockwise or clockwise turning of the body. 

Let the irregular outline (fig. 174) represent the projection 
of the moving body on the plane of the 
motion, AB a fixed line of the projection, 
OX a fixed reference line, and let 6 denote 
the angle XOA, it being regarded as posi- 
tive or negative according as OX, when 
turned about toward AB, turns counter- 
clockwise or clockwise. If X and 6 2 denote 
initial and final values of 6 corresponding to 
any motion of the body, then the 

angular displacement ==0 2 — d 1 = Ad. 

219. Angular Velocity and Angular Acceleration. — If a body 
has a plane motion, its angular velocity is the time-rate at which 
its displacement occurs, and its angular acceleration is the time- 
rate at which its angular velocity changes. 

These definitions are precisely similar to those of the angular 
velocity and acceleration of a rotation (arts. 210 and 213) ; hence 
the expressions, units, and rules of signs given in those articles 
and the following hold for any plane motion. Rewriting the 
expressions, 

co = d#/dt , and a = dco/dt = d 2 0/dt 2 , 

co and a denoting angular velocity and acceleration of the mov- 
ing body respectively. 




§ III.] PLANE MOTION. 209 

EXAMPLE. 

1. Determine the angular velocity and acceleration of the 
connecting-rod of a steam-engine, assuming the angular veloc- 
ity of the crank to be constant. 

Solution: Let OP and CP (fig. 175) be the crank and con- 
necting-rod respectively, their lengths being c and r, and let the 




Fig. 175. 

angles XOP and XCP be denoted by 6 and <f> (measured from 
the horizontal line and counter-clockwise positive) ; then for all 
positions 

rsin<f> = csmd, or cp = sm~ 1 ( -sin#). 

Let co and a denote the angular velocity and acceleration of the 
rod respectively, and co c the angular velocity of the crank. Then, 
since 

to = d<j>/dt, a=d 2 cf>/dt 2 , co c = dd/dt, and d 2 0/dt 2 = o, 

cos d 
(r 2 /c 2 -sin 2 0)* 



« J = /„2/^2 ZTmT ^ • • C 1 ) 



A (r 2 /c 2 -i)s md 2 , . 

and a = - ' 2 . 2 a co c 2 (2) 

(ryr-sm 2 Op 

2. Take r/c equal to 4 and compute the values of the coeffi- 
cients of co c and to 2 in eqs. (1) and (2) of the preceding solution, 
when = o°, 30 , 6o°, 90 , 120 , 150 , and 180 . Plot those 
values, thus showing how co and a vary during a stroke. 

220. Velocity and Acceleration of any Point of the Body. — 
Let P and P' (fig. 176) be two points of the moving body and O 
a point without, the three being in a plane parallel to that of the 
motion. According to art. 196 the velocity of P relative to 
equals the vector sum of the velocities of P relative to P' and P' 
relative to 0. Let the last velocity be v' directed as shown, and 
let r denote the distance PP' and co the angular velocity of the 



2IO 



MOTION OF A RIGID BODY. 



[Chap. IX. 



body at the instant considered. Since the motion of P relative 
to P' is circular, the velocity of P relative to P r equals rco (art. 
216), and its direction is perpendicular to PP' as shown. The 
velocity of P relative to then is the vector sum of v' and rco. 





Fig. 176. 

Similarly, the acceleration of P relative to equals the vec- 
tor sum of the accelerations of P relative to P' and P' relative 
to 0. Let the last acceleration be a' directed as shown, and let 
a denote the angular acceleration of the body at the instant con- 
sidered. The path of P relative to P' being a circle, the tan- 
gential and normal components of the acceleration of P relative 
to P' are respectively ra and rco 2 (art. 216) directed as shown, 
and the acceleration of P relative to O is the vector sum of a', 
ra, and rco 2 . 

Proposition. — The components of the velocities of any two 
points of a body having a plane motion * along the line join- 
ing them are equal and agree in sense. 

Proof: Let P and P' (fig. 176a) be two points in the plane of 
the motion. Since the velocity of P (v) is the resultant of rco 
and v' and rcois perpendicular to PP', the component of v along 
PP' is the same as that of v f along that line ; but 7/ is the veloc- 
ity of P' , hence, etc. 

It will be noticed that the proof is not general, the points 
being in the plane of the motion; this is the case to which the 
proposition is applied later. 



* Really true in any motion of a rigid body. 



§111.] PLANE MOTION. 2I1 



EXAMPLES. 

i . A wheel rolls uniformly and makes one turn every second. 
What is its angular velocity? 

Solution: Any line of the body parallel to the plane of the 
wheel describes an angle of 360 each second; hence the angular 
velocity is 360 deg.-per-sec. 

2 . Determine the velocity of any point on the rim of a wheel 
of radius r whose angular velocity is co rad. -per-unit time. 

Solution: We use the foregoing principles, choosing the 
centre of the wheel as P'. Let 6 denote the angle described by 

the radius PP' after any origin of time; p 

then the distance (s) travelled by P' in /^^^^>Xrw 

that time is given by 5 = rd. Hence / \ v ^^\ 

ds/dt = r dd/dt, or the velocity of P' at / r w /A 

any instant equals r times the angular / /' \ 

velocity of the wheel at that instant (see I Sw 

fig- T 77)- Relative to P f , the selected \ 

point P on the rim describes a circle and \ 

the velocity of that point relative to P' \^ 

is rco (art. 216), its direction being that /^^mMMMW 

of the tangent to the circle at P. Hence FlG - z 77- 

the absolute velocity of P (v) is represented by the diagonal Pp 

of the parallelogram on the two velocities rco. 

3. From the result of the preceding solution, determine the 
velocities of the highest and lowest points on the rim of the 
wheel. 

221. Plane Motion Regarded as a Combined Translation and 
Rotation. — Imagine the velocity of each point, P lt P 2t etc., of 
the moving body to be resolved into two components, one of 
which is the same as the velocity of any particular point P' of 
the body (fig. 178a). It follows from the preceding article that 
the other component is perpendicular to the line joining the 
point with P' and is equal to the product of the length of the 
line and the angular velocity of the body. 

Also imagine the acceleration of each point resolved into two 
components, one of which is the same as the acceleration of P' 
(fig. 1786). It follows from the preceding article that the other 



MOTION OF A RIGID BODY. 



[Chap. IX. 



component can be resolved into two components directed along 
and perpendicular to the line joining the point with P' ', they 
being equal to the products of the length of the line and the 
square of the angular velocity and angular acceleration respec- 
tively. 




Fig. 178. 

Now if the points of the body had the first sets of component 
velocities and accelerations only, the motion of the body would 
be a translation, the velocity and acceleration of which would be 
like those of the chosen point P' '. And if the particles had the 
second sets of component velocities and accelerations only, the 
motion of the body would be a rotation about the line through 
the chosen point and perpendicular to the plane of the motion, 
the angular velocity and acceleration of the rotation equal- 
ing respectively the angular velocity and acceleration of the 
actual motion. The two motions are hence regarded as compo- 
nents of the actual motion. 

222. Instantaneous Axis (of no Velocity). — Proposition. — If 

a body has a plane motion which 
is not translatory, then at each 
instant there is a line in it or in 
its extension all points of which 
have no velocity. 

Proof: Let P and P' (fig. 179) 
be any two points in the plane 
of the motion, and draw lines 
PO and P'O perpendicular to the directions of the velocities of 




Fig. 179. 



§111.] PLANE MOTION. 213 

P and P' respectively. The velocities of points on PO and P'O 
have no components along those lines respectively (prop., art. 
220), and since is on both lines its velocity has no components 
along PO and P'O, and hence the velocity must be zero. All 
points on the line through perpendicular to the plane of the 
motion have the same velocity as that of 0, i.e., zero. 

If the motion is translatory, the velocities of P and P' have 
the same direction and the point is "at infinity," but no points 
of a finite extension of the body have a zero velocity. 

Definitions. — The line of a moving body, all points of which 
have at a certain instant no velocity, is called the instantaneous 
axis of the motion at that instant. The intersection of the 
instantaneous axis with the plane of the motion is called the 
instantaneous centre. 

In general the instantaneous centre moves about in the body 
and in space. Its path in the body (i.e., the path relative to 
axes fixed in the body) is called body centrode, and its path in 
space (i.e., that relative to axes fixed in space, or in the earth) 
is called space centrode. 

It follows from the solution of exs. 2 and 3, art. 220, that the 
velocity of the lowest point of a rolling wheel is zero ; hence that 
point is the instantaneous centre. The body centrode is the 
circumference of the wheel, and the space centrode is the line on 
the plane surface along which the wheel rolls. 

223. Instantaneous Rotation. — Let P and Q denote two 
points of a body in a plane of motion, the latter being such 
that at some instant during the motion it is an instantaneous 
centre. At all times the motion of P relative to Q is circular 
and its velocity relative to Q at any instant equals the product 
of the distance between P and-Q (r) and the angular velocity of 
the body at that instant. 

Consider now the state of the motion when Q is the instan- 
taneous centre, calling the angular velocity of the body at that 
instant co. Since the absolute velocity of Q is zero, the absolute 
velocity of P (v) equals its velocity relative to Q, or 

v = toj (1) 

Hence the absolute velocity of any point of a mov- 



2I 4 



MOTION OF A RIGID BODY. 



[Chap. IX. 



ing body at any instant equals the product of the 
distance of the point from the instantaneous axis 
and the angular velocity of the body at that in- 
stant. 
Since in a rotation about a fixed axis the velocity of any 
point of the body is also proportional to its distance from the 
axis of rotation, the state of a plane motion at any instant is 
described as an instantaneous rotation about the instantaneous 
axis. 

EXAMPLES. 

i . Show how to find the instantaneous centre of the connect- 
ing-rod of a steam-engine in any position. 

Solution: The directions of the velocities of the ends of the 
rod are known for all positions; that of C (fig. 180) is OC and 
that of P is the same as that of the tangent to the crank-pin 
circle at P. Hence to find the instantaneous centre, draw per- 
pendiculars to the directions of these velocities at C and P re- 
spectively ; their intersection is the instantaneous centre of the 
rod for the position represented. 

2. Where is the instantaneous centre when the crank is hor- 
izontal? When vertical? What can you say about the state 
of the motion of the rod in these cases ? 

3. Show how to find the angular velocity of the connecting- 




Fig. 180. 



rod and the velocity of the cross-head C (fig. 180) in terms of 
the velocity of the crank-pin P for any position. 



§111.] PLANE MOTION. 215 

Solution : Let c denote the length of the crank ; 

v lt " " velocity of the crank-pin ; 
v 2 , " " " " cross-head; 

co f " angular velocity of the connecting- 

rod. 



According to eq. (1), co = vJIP and 

v 2 =IC'(o = (IC/IP)v v 

If it is desired to draw a velocity-space curve for the motion 
of the cross-head (as the dotted curve of the figure) when the 
crank-pin velocity is constant, the following simple construc- 
tions may be employed : Draw a vertical diameter of the crank- 
pin circle and mark its intersection with the connecting-rod 
(extended if necessary) C'\ then to the same scale by which OP 
represents the crank-pin velocity OC represents the cross-head 
velocity. For, from the figure, 

IC/TP = OC'/c, hence v 2 /v x = OC'/c. 

4. A sheet of paper is caused to slide on a draughting-board 
so that two points (P and Q) of the paper move along two lines 
(OX and OY) on the board. Show how to find the instanta- 
neous centre of the motion for any position of the paper. 

5. The velocity of one point of the paper of the preceding 
example being given, show how to find the velocity of any other 
point. 

EXERCISE. 

Take a sheet of paper and move it as described in ex. 4, the 
lines OX and OY being taken at right angles to each other, and 
determine the instantaneous centre for several positions of the 
paper. As each instantaneous centre is determined, mark it by 
pricking a hole through the paper and into the board, and join 
the holes in the paper and those in the board by smooth 
curves. These curves are the body and space centrodes 
respectively. 

Now cut the paper along the centrode and replace it on the 



216 MOTION OF A RIGID BODY. [Chap. IX. 

board in one of its positions (P and Q falling on OX and OY 
respectively). Then move the paper so that the body centrode 
rolls on the space centrode. If carefully done (a template fitted 
to the space centrode helps to get the rolling motion), it will be 
noticed that P and Q move along OX and OY; hence the actual 
motion is a rolling of the body centrode over the space centrode. 
It can be shown that any plane motion is equivalent to such a 
rolling. 



KINETICS. 



CHAPTER X. 

MOTION OF A PARTICLE (RESUMED) AND OF A SYSTEM OF 

PARTICLES. 

§ I. Mass and Mass-Centre. 

224. Quantity of Matter. — It must be confessed that the 
usual definition of mass (art. 12) needs explanation. The ques- 
tion at once arises, How shall matter be measured? If this is 
answered, then the meaning of "quantity of matter" is clear. 

"As long as we have to do with bodies of the same exact kind, 
there is no difficulty in understanding how the quantity of mat- 
ter is to be measured. If equal quantities of the substance 
produce equal effects of any kind, we may employ these effects 
as measures of the quantity of the substance. 

"For instance, if we are dealing with sulphuric acid of uni- 
form strength, we may estimate the quantity of a given portion 
of it in several ways. We may weigh it, we may pour it into a 
graduated vessel and so measure its volume, or we may ascertain 
how much of a standard solution of potash it will neutralize. 

"We might use the same methods to estimate a quantity of 
nitric acid if we were dealing only with nitric acid; but if we 
wished to compare a quantity of nitric acid with a quantity of 
sulphuric acid we should obtain different results by weighing, by 
measuring, and by testing with an alkaline solution." * 

Now these methods are not equally appropriate, and indeed 
that of titration cannot be applied to all bodies, that of measur- 
ing would lead us to say that the amount of gas in a tight rubber 
bag could be changed by simply squeezing (changing the volume 

* Quoted from Maxwell's "Matter and Motion." 

217 



218 



MOTION OF A PARTICLE. 



[Chap. X. 



of the bag), which is absurd. It will be shown that the' method 
of weighing is an appropriate one. 

Any appropriate method must be based on a common prop- 
erty of matter. Now all matter is inert, i.e., force must be 
applied to any body to change its velocity, and this property 
(inertia) is the basis of the fundamental method of determining 
quantity of matter. Not only is this method (explained in de- 
tail later) employed in mechanics, but also sometimes in ordi- 
nary affairs. Thus, suppose that we wish to ascertain whether 
a barrel lying upon a floor is full or empty ; we push it and con- 
clude that it is full or empty according as a large or small force 
is required to roll it, i.e., to "overcome the inertia." 

Along this line we can determine the relative amounts of 
matter in two bodies if we have a means of measuring forces. 
Thus, calling the two bodies A and B, place each upon a light 
and easy-running carriage (fig. 181) and connect these by means 




Fig. 181 



of cords to spring-balances which rest upon and are fastened to 
a third carriage as shown. If this last carriage is pulled to the 
right, the others follow and the spring-balances measure the pulls 
required to move the smaller carriages and their loads. Clearly 
it will be in accord with the crude test applied to the barrels to 
say that the quantities of matter in A and B (neglecting that in 
the carriages) are as the forces applied to them, as read from 
the spring-balances. It will also be in accord with a comparison 
of quantities of matter in bodies of the same kind by the method 
of volumes ; for suppose that A and B are of the same kind and 
that the volume of A is n times that of B, we would say at once 
that there is n times as much matter in A as in B. The same 



1 1.] MASS AND MASS-CENTRE. 219 

ratio would be arrived at by the inertia method, i.e., from the 
readings of the spring-balances. 

We now make definite our notions about quantity of matter 
as just expressed by means of the following 

Definition. — -The quantities of matter in bodies are propor- 
tional to the forces required to give them equal accelerations. 

If we should adopt as a standard, or unit, the amount of 
matter in any particular body, we could determine the quantity 
in any other body in terms of this unit (ideally at least) by plac- 
ing it and the standard body on the two smaller carriages and 
then measuring the forces required to give them the equal accel- 
erations. If the force on the body in question is n times that 
on the standard, the quantity of matter in the body would be n. 

Of course this scheme of measurement is not put forth as a 
practical one, but rather as a help to understand the meaning of 
mass. The practical method of measuring quantity of matter is 
by weighing, which is (as explained in art. 226) precisely similar 
in principle to that just described. 

225. Mass. — Definition (repeated from art. 12). — By mass of 
a body is meant the quantity of matter in it. The word is 
merely an abbreviation for " quantity of matter." 

Units of Mass. ^-There are many units of mass in use; they 
may be grouped into two classes: 

(a) Absolute Units; so called to express the fact that their 
magnitudes are independent of locality. Two of these, the 
pound and the kilogram, are described in art. 12. Their rela- 
tion is as follows: 

1 pound = 0.4536 kilograms, 
or 1 kilogram = 2.205 pounds. 

(b) Gravitation Units; so called because their magnitudes 
vary with locality precisely as acceleration due to gravity varies. 
Two of these units are described in art. 233. 

226. Practical Determination of Mass. — As is well known, 
bodies falling at the same place in vacuum move with equal 
accelerations, i.e., the forces of gravity upon bodies (their weights) 
at the same place accelerate them equally. It follows from this 
fact and the definitions of arts. 224 and 225 that 



220 MOTION OF A PARTICLE. [Chap. X. 

the masses of bodies are proportional 
to their weights at the same place. 
Hence to determine the mass of a body, i.e., to determine the 
ratio of its mass to that of a standard (pound for example), we 
determine the ratio of the weights of the body and the standard. 
If this latter ratio is n, then the former is n also, and the mass 
of the body is n times that of the standard (or n pounds). 

227. Moment of Mass. — The product of the mass of a particle 
and its ordinate with respect to any plane is called the moment 
of the mass of the particle with respect to that plane. An ordi- 
nate is regarded as positive or negative according as the parti- 
cle is on one side of the plane or the other ; hence a moment has 
the same sign as the corresponding ordinate. 

By moment of the mass of a system of particles is meant the 
algebraic sum of the moments of the masses of its particles. 
Thus, let the particles of a system be referred to a set of rect- 
angular axes, and denote the coordinates of the particles by 
(x v y t , z t ), (x 2 , y 2 , z 2 ), etc., and their masses by m v m 2 , etc. Then 
the moments of the mass of the system with respect to the y-z, 
z-x, and x-y planes are respectively 

m 1 x 1 +m 2 x 2 + . . . = Imx; 
m 1 y 1 + m 2 y 2 -f- . . . = 2 my ; 
m 1 z 1 + m 2 z 2 +.-...= Imz. 

228. Mass-Centre Defined. — It is obvious that the mass of a 
system may be multiplied by some distance (positive or nega- 
tive) such that the product equals the moment of the mass with 
respect to a given plane. Thus, let M be the mass of a system 
and x,y, and z such multipliers that 

Mx = Imx, My = Imy, Mz = Imz. . . . (1) 

The point whose coordinates are x, y, and z is called the mass- 
centre of the system. The formulas for the coordinates of the 
mass-centre are therefore 

_ _ Imx _ _ 2my _ _ Jmz ( . 



§11.] MOTION OF A PARTICLE. 221 

229. Relation Between Mass-Centre and Centre of Gravity* — 

Since the masses of the particles of a system are proportional 
to their weights, we may substitute for the mass terms in the 
equations (2) the corresponding weights, i.e., 

_ _ Iwx _ _ Iwy _ _ Iwz 

X== ~W> y ~~W' z ~lv~' 

w denoting the weight of any particle whose coordinates are x, y . 
and z, and W the weight of the system. But these values of ~x, 
y, and z are identical with those for the x, y, and z coordinates 
of the centre of gravity of the system (see art. 64); hence the 
mass-centre and centre of gravity of a system of particles (as 
herein defined) are coincident. 



§ II. Motion of a Particle. 

230. Laws of Motion. — (1) When no force is exerted upon a 
particle it remains at rest or continues to move uniformly in a 
straight line. 

(2) When a force is exerted upon a particle it is accelerated; 
the direction of the acceleration is the same as that of the force , 
and its magnitude is proportional to the force directly and to the 
mass of the particle inversely. 

(3) When one particle exerts a force upon another the latter 
exerts one on the former, and the two forces are equal, collinear, 
and opposite.* 

These laws are inductions from observation and experiment, 
made not of course on particles, but on bodies of ordinary size. 
We do not attempt a full discussion of the experience leading to 
the laws, but limit ourselves to a brief statement. Really the 
best evidence of the correctness of the laws are the many agree- 
ments noted between observed results and those calculated from 
the laws, and the fact that the laws are not known to be in dis- 
agreement with any phenomenon. 

( 1 ) We know of no body which is free from the influence of 

* These are essentially Newton's Laws of Motion. The form of state- 
ment here given, however, differs from that in which they were originally 
announced (1687). 



222 MOTION OF A PARTICLE. [Chap. X. 

all others, i.e., a body not acted upon by force, and so no direct 
observation or experiment leads to this law. But all have per- 
haps noticed that a small body, if projected along a sheet of ice, 
moves in a straight path and continues to move for a consider- 
able period; also that the smoother the ice the longer will the 
body move, i.e., the smaller is its retardation or the more nearly 
is its motion uniform. The retardation is rightly ascribed to the 
frictional resistance offered by the ice ; and it is a fair inference 
that if that resistance were zero, the retardation would also be 
zero and the motion would be uniform. 

(2) The second law may be roughly verified by the means of 
the apparatus represented in fig. 181. Thus place a body whose 
mass is known on one of the small carriages and another on the 
hook C. Then let the system move, measure the acceleration of 
A and the pull recorded by the spring-balance. If this is re- 
peated with other bodies on C, it will be found that the acceler- 
ations are roughly proportional to the pulls. Also place bodies 
of unequal mass successively on A and cause the carriages to 
roll so that the spring-balance reading is the same for the differ- 
ent bodies, measuring the acceleration of each motion. It will 
be found that the accelerations are roughly inversely propor- 
tional to the masses of the bodies. The lack of exact propor- 
tionality may properly be ascribed to the neglect of friction and 
the mass of the carriage A, and to unavoidable experimental 
errors involved in such an apparatus. 

(3) It is well known that a magnet exerts a force on a piece 
of soft iron in its proximity ; that the iron also exerts a force on 
the magnet, may be proven as follows: Place the magnet and 
piece of iron in small vessels ; then float them so loaded in water. 
If they are not too far apart they will be observed to move 
toward each other ; hence not only does the magnet attract the 
soft iron, but the iron attracts (exerts a force) on the magnet. 

If two spring-balances be laid on a table, the hook of one 
engaging the hook of the other, and then they be pulled apart 
by their rings, each will register the pull exerted upon it by the. 
other. If the balances are accurate, it will be found that the 
amounts registered are equal; hence the forces exerted by the 
hooks on each other are equal. 



§11.] MOTION OF A PARTICLE. 223 

231. Quantitative Expression of the Second Law of Motion. — 

Let m' and m" denote the masses of two particles and a' and a" 
the accelerations given them by two forces F'and F" respectively. 
The second law of motion asserts that 

a':a"::F'/m':F"/m" i or F'/m'a'=.F"/m"a". 

It follows from this equation that the ratio of the force acting 
on a particle to the product of the mass and the acceleration of 
the particle produced by the force is constant, i.e., in the form 
of an equation, 

F/ma = k, or F = k-ma, (1) 

k being the value of the constant ratio. 

The numerical value of k depends upon the units used for 
expressing value of force, mass, and acceleration. It is con- 
venient, but not necessary, so to choose these units that k 
becomes 1. A system of units so chosen is called a kinetic 
system. 

232. Kinetic System of Units. — It follows from the last equa- 
tion that in any kinetic system the unit force acting upon the 
unit mass produces unit acceleration (i.e., unit velocity in unit 
time). Evidently any three of the units involved in a kinetic 
system (units of force, mass, length, and time) may be chosen 
arbitrarily, but the fourth must be such as to satisfy the re- 
quirement just stated. The unit of force or of mass is the one 
selected as the derived or fourth unit. There are two classes of 
kinetic systems. 

(a) Absolute Kinetic Systems. — Units of length, mass, and 
time are arbitrarily selected, the unit of force being derived. 

Centimetre- Gram-Second (cg.s.) System. — The units of 
length, mass, and time are the centimetre, the gram (one-thou- 
sandth of a kilogram, see art. 12), and the second respectively. 
The corresponding unit of force, i.e., the force which acting on 
a gram mass for one second gives it a velocity of one centimetre 
per second, is called a dyne. This is the system now univer- 
sally used in scientific work and literature. 

Foot-Pound-Second (f.p.s.) System. — The units of length, 
mass, and time are the centimetre, the pound (see art. 12), and 



224 MOTION OF A PARTICLE. [Chap. X. 

the second respectively. The corresponding unit of force, i.e. , the 
force which acting on a pound mass for one second gives it a 
velocity of one foot per second, is called a poundal. This sys- 
tem has never met with favor; accordingly, it is not used herein, 
but is mentioned and explained because of the relations which 
it bears to other important systems. 

(b) Gravitation Kinetic Systems (also called Engineers' Sys- 
tems). — Units of length, force, and time are arbitrarily selected 
and the unit of mass is derived. The units of force in these 
systems are weights of absolute units of mass, i.e., they depend 
upon the force of gravity and are gravitation units (see art. 13) 
The units of length and time being independent of place, the units 
of mass in these systems must vary just as the units of force 
vary, and are hence properly called gravitation units of mass. 

Foot-Pound (force) -Second System. — The units of length, 
force, and time are the foot, the pound (see art. 13), and the 
second respectively. The corresponding unit of mass, i.e., one in 
which a pound force would produce in one second a velocity 
of one foot-per-second, is called herein a geepound* 

Metre-Kilogram (force) -Second System. — The units of length, 
force, and time are the metre, the kilogram (see art. 13), and the 
second respectively. The corresponding unit of mass, i.e., one 
in which a kilogram force would produce in one second a veloc- 
ity of one metre-per-second, is called herein a geekilogram* 

233. Relations Between Force Units and Between Mass Units. 
— The dyne and the poundal might ideally at least be ' 'preserved " 
as follows : Place one gram (or one pound) on one of the small 
carriages of fig. 181, and then make the large carriage move to 
the right with an acceleration of one cm.-per-sec.-per-sec. (or 
one ft.-per-sec.-per-sec.) and note the reading of the spring- 
balance. Assuming the mass of the small carriage compared to 

* This is a new term, and the student should remember that fact, for 
it is not now and perhaps never will be current. The unit is usually 
called "engineers' unit of mass," an appellation which is on a par with 
"wood-choppers' unit of volume" (the cord). Other terms have been 
proposed, but they have never been adopted. The author is confident 
that names for gravitation units of mass are convenient and that their use 
tends to clearness. 



§11.] MOTION OF A PARTICLE. 225 

that of its load, and its frictional resistance compared with the 
balance reading to be negligible, the force causing the stretch of 
the spring is one dyne (or one poundal). 

Instead of employing the foregoing method, we make use of 
measurements made on the acceleration due to gravity, thus 
comparing the dyne and poundal with standards or units of 
weight. A force equal to the weight of a gram (or a pound) act- 
ing on one gram (or one pound), as in a falling body, produces 
an acceleration of approximately 981 cm.-per-sec.-per-sec. (or 
32.2 ft.-per-sec.-per-sec); hence a force equal to 1/9 81 of the 
weight of a gram (or 1/32.2 of the weight of a pound) would 
produce in one gram (or one pound) an acceleration of one cm.- 
per-sec.-per-sec. (or one ft.-per-sec.-per-sec). Now by defini- 
tion the forces producing in one gram and in one pound acceler- 
ations of one cm.-per-sec.-per-sec. and one ft.-per-sec.-per-sec. 
are the dyne and poundal respectively; therefore 

one dyne = 1/98 1 ± gram (force), 

one poundal = 1/32.2 ± pound (force).* 

The "geepound" and the "geekilogram" might also be deter- 
mined by means of the apparatus represented in fig. 181. Thus 
adjust a load on the smaller carriage so that the spring-balance 
will read one pound (or one kilogram) when the larger carriage 
is drawn to the right with an acceleration of one ft.-per-sec.-per- 
sec. (or one m.-per-sec.-per-sec). If the mass and frictional 
resistance of the small carriage is negligible, the mass of the load 
is one geepound (or one geekilogram). 

A better determination of these units of mass can be made 
by means of experiments on the acceleration due to gravity. A 
one-pound (or a one-kilogram) force produces in a one-pound 
(or a one-kilogram) mass an acceleration of approximately 32.2 
ft.-per-sec.-per-sec. (or 9.81 m.-per-sec.-per-sec); hence a one- 
pound (or a one-kilogram) force would produce in 32.2 pounds 
(or 9.81 kilograms) mass an acceleration of 1 ft.-per-sec per-sec. 
(or one m.-per-sec.-per-sec). Now by definition, two masses 
which under the action of forces of one pound and one kilogram 



* One gram and one pound (force) being the weight of one gram and 
one pound (mass) respectively, 



226 MOTION OF A PARTICLE. [Chap. X. 

receive accelerations of one ft.-per-sec.-per-sec. and one m.-per- 
sec.-per-sec. are the geepound and the geekilogram respectively. 
Therefore 

i geepound = 32.2 ± pounds, 

1 geekilogram = 9.8 1 ± kilograms. 

234. Relation Between the Mass and the Weight of a Body. — 

This relation is implicitly given in the preceding article ; we will 
now state it definitely. Let W denote the weight of a body, 
m its mass, and g its acceleration due to gravity, the three quan- 
tities being expressed in units of any one kinetic system. When 
this body falls, its acceleration (g) is due to its weight (W); 
hence the equation of the motion is (see art. 231) 

W = mg, or m = W/g, 

and this gives the relation between the weight and the mass of 
any body when they (W, m, and g) are expressed in units of any 
one kinetic system. Thus for any body 

W (in dynes) =m (in grams) X981 ± , 

W (in poundals) = m (in pounds) X 3 2 . 2 ± , 
m (in geepounds) = W (in pounds) -s- 3 2 .2 ± , 
m (in geekilograms) = W (in kilograms) 4-9.81 ± 

The weight and mass of a body are also numerically the 
same if expressed in certain units, more definitely if the unit 
weight is the weight of the unit mass (see art. 15). Thus the 
weight of a barrel of flour is 196 pounds and its mass is also 196 
pounds. 

EXAMPLE. 

Express the mass of a cubic foot of water in pounds, gee- 
pounds, kilograms, and geekilograms. 

235. Acceleration of a Particle Acted Upon by Several Forces. 

— The acceleration of a particle acted upon by several forces 
may be determined in several ways : 

(a) By the methods of " Statics " determine the resultant of 
the forces and then compute the acceleration due to this resultant. 
This acceleration is identical with that due to the actual forces 
applied to the particle, for by definition the resultant of any 
number of forces is equivalent to them in producing motion. 



§ II.] MOTION OF A PARTICLE. 227 

{b) Compute the acceleration of the particle due to each 
force acting alone and add these accelerations vectorially. The 
vector sum represents the actual acceleration due to the com- 
bined action of the forces. The correctness of this method may 
be proved from the first; we give the proof for the case of two 
forces: Let m be the mass of the particle, F x an^. F 2 the applied 
forces, a the acceleration due to their combined action, R their 
resultant, and a x and a 2 the accelerations due to F t and F 2 acting 
singly. Let AO and BO (fig. 182) represent F x and F 2 respect- 



er 




Fig. 182. 

ively; then the diagonal of the parallelogram OABC repre- 
sents R (art. 20), and the continuation Oc of that diagonal 
represents a if Oc=R/m (art. 231). Also, Oa and Ob represent 
a t and a 2 respectively if 

Oa = F 1 /m and Ob = F 2 /m. 

It remains to show that Oc is the vector sum of Oa and Ob. To 
do this, join a and c and b and c, and show that Oabc is a par- 
allelogram. 

236. Equations of Motion of a Particle. — Let m, a, and R de- 
note the mass and acceleration of a particle and the resultant of 
the forces applied to it respectively. Then, as explained in the 
preceding article, 

R = ma, (1) 

and a and R have the same direction. 

Let F f , F", etc., denote the forces applied to the particle and 
a, /?, and y the angles which R and a make with a set of coordi- 
nate axes x, y, and z. Also let IF X , IF y , and IF Z denote the 
algebraic sums of the x, y, and z components of F' ,F" , etc., and 
a x , a y , and a z the x, y, and z components of a. Now 



228 MOTION OF A SYSTEM OF PARTICLES. [Chap. X. 

R cos a = ma cos a, i^cos /? = ma cos /?, and R cos y = ma cos 7- ; 

and since i? cos a = IF X , R cos p = IF y , R cos y = IF g , 
and acosa=a 2 , acos/3 = a y , acos7' = a 2 , 

i , F x = ma x , i , F 2/ =ma 2/ , i'F 2 =ma 2 . ... (2) 

Let <f> denote the angle between the action line of R and the 
tangent to the path of the particle. Then 

R cos <j> = ma cos <f> and R sin </> = ma sin <£, 

or 2T, = ma, and i , F n = ma n (3) 

XF, and IF n denoting the algebraic sums of the tangential and 
normal components of the forces, and a t and a n the tangential 
and normal acceleration of the particle. 

§ III. Motion of a System of Particles. 

237. Definitions. — Any number of particles collectively con- 
sidered are called a system of particles. If the distances between 
the particles remain invariable, the system is called a rigid one 
or a rigid body. 

Among the forces exerted upon the system of particles, some 
may be exerted by particles not belonging to the system. Such 
forces have already been named external forces (art. 113), and all 
such forces the external system. A force exerted by a particle 
upon another of the same system has been named an internal 
force (art. 113), and all such forces the internal system. 

According to the third law of motion (art. 230), if one particle 
of a system exerts a force upon another, the second also exerts 
one upon the first, and these two forces are equal, opposite, and 
collinear. Hence the internal forces of a system of particles 
occur in pairs, the forces of each pair being equal, opposite, and 
collinear. 

By effective force for a particle of a system is meant the re- 
sultant of all the forces (external and internal) acting on that 
particle. The effective forces for all the particles of a system 
are called the effective system. If m and a denote the mass and 



§m]. MOTION OF A SYSTEM OF PARTICLES. 229 

acceleration of any particle of a system, then (art. 236) for that 
particle the 

effective force = ma. 

238. D'Alembert's Principle. — Since the effective system con- 
sists of the external and the internal forces acting upon the par- 
ticles of a system, the resultant of the effective forces is identical 
with that of the external and the internal forces. Now the 
resultant of the internal forces is zero since they occur in pairs, 
the forces of each being equal, opposite, and collinear; hence 

Proposition I. — For any system of particles the resultants 
of the effective and the external forces are identical. Obviously 
this proposition is equivalent to 

Proposition II. — For any system of particles the effective 
forces reversed * and the external forces together are in equilib- 
rium. This proposition is known as D'Alembert's Principle, 
after him who first announced it (1742) for rigid bodies. 

These propositions are not fundamental, being deducible 
from the laws of motion (art. 230), but they express an impor- 
tant relation in convenient forms. 

239. Component of an Effective System Along Any Line. — 
Let the line be taken as an x axis, and as before let m lf m 2 , etc., 
denote the masses of the particles and a x ', a x " , etc., the x com- 
ponents of their accelerations. Then the algebraic sum of the 
x components of the effective forces for the particles is 

m 1 a x '+m 1 a x " + . . . 

This sum equals the product of the mass of the system (Im) and 
the x component of the acceleration of its mass-centre (a z ). 
For, according to art. 228, 

m 1 x 1 +m 2 x 2 + . . . = Im-x % (1) 

hence m 1 dx 1 /dt + m 2 dx 2 /dt+ . . . = Im-dx/dt, ... (2) 

and m x d 2 xJdt 2 -\-m 2 d 2 x 2 /dt 2 + . . . = Im-d 2 x/dt 2 , . . (3) 

or m x a x ' -\-m x a x " + . • • = 2m-a x (4) 

240. Motion of the Mass-Centre of any System of Particles. — 

According to Prop. I, art. 238, the algebraic sums of the com- 

* Often called the "reversed effective system." 



230 MOTION OF A SYSTEM OF PARTICLES. [Chap. X. 

ponents of the external and effective forces for any system of 
particles along any line are equal. Hence, if IF X , IF y , and 
1F Z denote the sums of the components of the external forces 
along an x, y, and z axis respectively, 

IF x = Imsi x> Z¥ y = Zm~k y , IF^m-a,. . . (i) 

These equations show that the acceleration of the mass-centre 
depends only on the values of the components of the external 
forces and that 

The acceleration of the mass-centre of a system is 
just like that of a particle, whose mass equals 
that of the system, acted on by forces equal and 
parallel to the external forces which are applied 
to the system. 

EXAMPLES. 

i. Show that if the air resistance were zero the acceleration 
of the mass-centre of any body thrown into the air in any way 
would equal g and be directed vertically downward. 

Solution: Let W denote the weight of the body, and refer 
the motion to a set of axes fixed in the earth, the y axis being 
vertical, its positive end being up. Then since the only force 
acting on the body during the motion is its weight, 

IF^=o = (W/g)a x , ZF y =-W = (W/g)a y , ZF z = o = (W/g)a z> 
or a x = a z = o, and a y = — g\ hence a = —g. 

2. The mass-centre of a "stationary" steam-engine when 
running is in general not a fixed point. Show that the reac- 
tion of its supports is not constant in amount. 

Solution : Let m and W denote the mass and weight respect- 
ively of the engine, and a the acceleration of its mass-centre. 
Also let R x , R y , and R z denote the x, y, and z components of 
the reaction of the supports, the y axis being taken vertical. 
Then eqs. (i) become 

R x = ma x , R y = — W + ma y , R z = ma g . 

3. What is the greatest acceleration which a locomotive can 
give a train? 



§111.] MOTION OF A SYSTEM OF PARTICLES. 231 

Solution: When the locomotive is pulling, the drivers tend 
to slip on the rails and the latter therefore exert on the former 
frictional forces directed forward. (These may be regarded as 
the forces which directly make the train move.) The rails exert 
on the other wheels of the train forces directed backward ; call 
the sum of their horizontal components R', and the resistance 
of the air R". If F denotes the sum of the frictional forces, m 
the mass of the train, and a its acceleration, 

F-R'-R" = ma, or a = (F -R' -R")/m. 

F is maximum when the drivers are about to slip, R" de- 
pends on the velocity, and is least when the velocity is zero, 
and R' does probably not depend much on the velocity and 
is practically independent of the acceleration. Hence the 
acceleration is greatest at low speeds if the drivers are about to 
slip. 

4. What can you say of the forces which a travelling crane, 
moving with an acceleration, exerts on its track? 

241. Moment of the Effective System About Any Axis. — Let 
the moment axis be taken as an x coordinate axis, and call the 
mass of any particle m, its acceleration a, its coordinates x, y, 
and z. The effective force for the particle is ma, and its x, y, 
and z components are ma x , ma y) and ma z respectively, and the 
moment of the force about the x axis equals the sum of the 
moments of its components (Prop. I, art. 28), i.e., 

ma z • y — ma y • z. 

The moment of the entire system about the x axis equals the 
sum of all such expressions as the above, or 

I{ma z y — ma y z). 

242. "Angular Motion" of a System of Particles. — According 
to Prop. I, art. 238, the moment sums of the external and effect- 
ive force systems about any axis are equal. Hence if 1M X , 
J My, and IM Z denote the moment sums of the external forces 
about the x, y, and z coordinate axes respectively, 



232 MOTION OF A SYSTEM OF PARTICLES. [Chap. X. 

IM X = I(ma. z y - ma^z) , 
IM y = I(ma. x z — ma g x) , 

IM Z = ^(ma^x-maj). 

It can be shown that these and eqs. (i) art. 240 completely 
determine the effect of a force acting upon a rigid body. Since 
the components and the moments of a force do not depend on its 
application point, the equations also do not ; hence the effect of a 
force on the motion of a rigid body does not depend on its appli- 
cation point. This is called the " principle of transmissibility." 



CHAPTER XL 
TRANSLATION OF A RIGID BODY (RESUMED). 

§ I. General Principles. 

243. Equations of Motion. — Let m denote the mass of a 
body, cl its acceleration, and a x , a y , and a z the x, y, and z com- 
ponents of a. Then, from the equations of motion of the mass- 
centre of any system having any motion (art. 240), 

i , F a .=ma a: , ZY y = mdLy, i , F 3 -ma z . . . . (1) 

IF Xy IF y , and IF Z denoting the algebraic sums of the x, y, 
and z components of the external forces. It is advantageous 
to select the coordinate axes in a certain way in special cases; 
thus if the motion is a plane one, two axes should be taken in 
the plane of the motion, for then the equations reduce to two in 
number ; and if the motion is rectilinear, one of the axes should 
be taken parallel to the direction of the motion, for then the 
number of equations reduces to one. 

It is shown in the next article that the resultant of the effect- 
ive forces for the particles of a translating body is a single force 
acting through the mass-centre in the direction of the accelera- 
tion, its magnitude being equal to the product of the mass of 
the body and its acceleration. According to D'Alembert's 
Principle, the resultants of the external and effective systems 
are identical; hence if the resultant of the external forces is 
denoted by R, 

R = ma, (2) 

and it acts through the mass-centre in the direction of the accel- 
eration. 

244. Resultant of the Effective System. — In a translation, 
the accelerations of all the particles of the moving body at each 
instant are alike in magnitude and in direction. Hence the 
effective system consists of forces having the same direction, 

233 



234 TRANSLATION OF A RIGID BODY. [Ch. XL 

and they are proportional to the masses of the corresponding 
particles; therefore the resultant of the effective system is a 
single force. 

Obviously the direction of the resultant at each instant is 
the same as that of the acceleration. The magnitude equals the 
sum of the separate effective forces, i.e., 

(dm 1 )a 1 + (dm) 2 a 2 + . . . =a I dm = ma, 

(dm) 1} (dm) 2 , etc., denoting the masses of the particles. The 
action line of the resultant passes through the mass-centre, as 
can be shown thus: The effective forces constitute a system of 
parallel forces with fixed application points ; hence they have a 
centre or centroid (art. 62). Let x, y, and z denote the coordi- 
nates of any particle with reference to a set of fixed axes, x , y , 
and z the coordinates of the centroid of the effective forces, and 
x, y, and T the coordinates of the mass-centre. Then (see art. 63 ) 



/ (dm • a)x I dm • x 



x n = = = x (see art. 228). 

ma m 

Similarly it can be shown that y =y and z = z; hence the cen- 
troid of the effective forces coincides with the mass-centre, and 
the resultant of the effective forces passes through the mass- 
centre as stated. 

§ II. Applications. 

245. General Method of Procedure. — In the following prob- 
lems the forces applied to a body are wholly or partially given 
and it is required to determine the acceleration, or else the 
acceleration is given and one or more forces are required. Such 
problems are solved by writing the equations of motion (or as 
many as necessary) and then solving them for the unknowns. 

If it is required to determine the motion completely, i.e., to 
compute the acceleration, velocity, and position at each instant 
of the motion, the acceleration is found first as just indicated 
and then the velocity and position may be found by methods 
explained in Chapters VII and VIII. 



§ ILJ APPLICATIONS. 235 



, EXAMPLES. 

i. If the body represented in fig. 183(a) weighs 50 lbs., the 
pull P is 40 lbs., the angle 6 is o, and the supporting surface is 



40 lbs. 

— ? — 




wwimnmimiii 



(a) (b) 

Fig. 183. 

smooth, compute the acceleration and the reaction of the sup- 
port. 

Solution: Let R denote the reaction of the plane; then the 
external forces acting on the body are as represented in fig. 
183(6). The mass of the body is 50/32.2 = 1.553 geepounds. 
Hence 

27^=40 = 1.553a*, 

IF v =R-$o = i.s$3a y , 

XF 2 = o = 1.5530^. 

From the third equation a z = o, and from the first a x = 25.76 
ft. /sec. 2 Obviously a y = o, hence the acceleration of the body 
is 25.76 ft. /sec. 2 in the plus x direction. Since a y equals zero, 
the second equation shows that R = $o lbs. 

2. Suppose that at a certain instant the mass-centre of the 
body of ex. 1 is at the origin, its velocity being 30 ft.-per-sec. in 
the plus x direction. Compute the velocity and position of the 
mass-centre two seconds later. 

3. Suppose that at a certain instant the mass-centre of the 
body of ex. 1 is at the origin and that the velocity of the body 
at that instant is 30 ft.-per-sec. in the plus z direction. Deter- 
mine the velocity and the position of the mass-centre two seconds 
later. 

4. Solve ex. 1, supposing that equals 20 . 

Ans. a = 24.2 ft. /sec. /sec. 

5. Suppose that P = 4o lbs. and d = o (fig. 183), that the sup- 
porting surface is rough, the frictional resistance being 10 lbs., 



236 



TRANSLATION OF A RIGID BODY. 



[Ch. XL 



and that at a certain instant the body is at rest. Determine the 
subsequent motion. 

6. Suppose that at a certain instant the body of ex. 5 is 
moving in the minus x direction with a velocity of 30 ft.-per- 
sec. Determine the subsequent motion. 

7. Suppose that the inclined plane (fig. 184a) is smooth, that 
the body upon it weighs 50 lbs., and that 
P equals 40 lbs. Determine the acceler- 
ation and the reaction of the plane. 

(°" Solution: Let R denote the reaction 
of the plane; then the external forces 
acting on the body are as shown in fig. 
1 84(6) . Taking coordinate axes as shown, 
the mass being 1.553 geepounds, 

(b) 2F X = — 40 + 50COS60 




i-553^ 



IF, 



R 
o = 



5osm6o° = i.553a 2/ , 
27^ = = 1.553^. 



Fig. 184. 



From the first and third respectively, 
9.66, a z = o, and obviously a y = o\ 



hence a = 9.66 ft. /sec. 2 in the negative x direction. 

From the second equation it follows that ^ = 43.3 lbs. 

8. Suppose that the plane in fig. 184(a) is rough and that the 
frictional resistance is 10 lbs. If at a given instant the velocity 
of the body is zero, determine the subsequent motion. 

9. Suppose that at the instant mentioned in the preceding 
example the velocity is 40 ft.-per-sec. down the plane. Deter- 
mine the subsequent motion. 

10. Fig. 185(a) represents an open box on a horizontal^ sur- 
face, the box containing a stone and subjected to a force P. Let 
the weights of the box and stone be 90 and 70 lbs. respectively, 
P 100 lbs., and suppose that the supporting surfaces are smooth. 
Compute the pressure between the stone and the rear end of the 
box. 

Solution: The external forces applied to box and stone 
together consist of the pull, the reaction of the horizontal sup- 
port and the weights (see fig. 1856). The masses of the two 



II.] 



APPLICATIONS. 



237 



bodies are 70/32.2 and 90/32.2, or 2.17 and 2.83 geepounds re- 
spectively. Evidently the acceleration a is in the direction of 
the 100-lb. force and its value is given by 

100 = 4.970, or a = 20.1 ft. /sec. /sec. 

The external forces acting on the box and stone respectively are 
shown in (c) and (d) , P' and R' denoting pressures between the 
stone and the end and bottom of the box respectively. Know- 




s 



R||90 



lbs. 
(b) 



R 1 DO lbs. 



fc£ 



(e) 



f«> 



100 lbs. 



200 lbs. 



" " " '"" >>»;<■<•>.;' ■ -■;,_ 



(e) 



Fig. 185. 



ing the acceleration of each body, the forces can be determined 
from the equations of motion of either; thus for the stone, 

P / = 2.i7X20.i=43.62 lbs. 

1 1 . Four bodies are connected by cords as shown in fig. i85(^) 
and are pulled along on a horizontal plane by a force P of 200 
lbs. Supposing the frictional resistance on each body to equal 
one-fourth its weight, compute the tension in each cord. 

12. In an elevator there are two boxes weighing 610 and 1000 
lbs. respectively, the lighter being on top of the other. What 
are the pressures on the bottoms of the boxes if (a) the elevator 
is started up with an acceleration of 4 ft. /sec. /sec? (b) If 
started down with the same acceleration? 

Ans. (a) 1 630 lbs. on the bottom of the lower box. 

13. If the elevator of the preceding example weighs 1600 
lbs., compute the tensions in the hoisting cable in the two cases 
mentioned. 

14. Fig. 186(a) represents two bodies, A and B, suspended 
from the ends of a cord which passes over a "smooth pulley " C. 
Let the weights of A and B be W 1 and W 2 (W 2 being the greater). 



2 3 8 



TRANSLATION OF A RIGID BODY. 



[Chap. XI. 



Assume that the cord and pulley are practically without 
mass and that the axle friction is zero; then the tension is the 
same at all sections of the cord. Compute the acceleration of 
the bodies and the tension. 



<kb 




%yi^y^<%%^^5^%^^^ 



a 



(a) 



W 



& 



Fig. 186. 



Solution: Let T denote the tension ; then the external forces 
applied to each body are as represented. Assuming the string 
to be inextensible, the accelerations of the two bodies are equal; 
we will denote them by a. Evidently the accelerations of A 
and B are respectively upward and downward; hence the re- 
sultant forces on A and B act up and down respectively, i.e., 
W t <T<W 2 . The masses of A and B being WJg and WJg 
respectively, the equations of motion for these bodies are 



T-Wi 



W, 



and Wn 



T = — 1 a. 

a 



Solving these equations for T and a, we find that 



W 2 -W, 



g and T 



2 W t W 2 



W t + W° w t +w 2 

15. Fig. 186(6) represents two bodies, A and B, connected 
by a cord passing over a smooth pulley, one hanging freely and 
the other supported by a horizontal surface. Let the weights 
of A and B be W x and W 2 respectively, and assume the surface 
to be smooth and that the tensions at different sections of the 
cord are equal. Compute the acceleration of the bodies and the 
tension. 

w x _ ^ w,w 2 



/ins. a = 



7 = 



W 1 + W 2 & ' W x + W 2 ' 

16. Solve the preceding ex., supposing that the W x — 6^ lbs., 
W 2 = o6 lbs., and that the horizontal surface is so rough that its 
frictional resistance on the body is 19.2 lbs. (Take £ = 32.) 



§ II.] APPLICATIONS. 239 

17. Suppose that a stone weighing 100 lbs. is placed on a 
rough horizontal board, the coefficient of static friction for the 
stone and board being 1/4. If the board is moved horizontally 
with an acceleration of 4 ft. /sec. /sec, will the stone remain on 
the board? 

18. Suppose that the stone of the preceding example is square 
in cross-section, 1 by 1 ft., and 8 ft. high, and that it stands so 
that two of its sides are parallel to the direction of the motion. 
How large an acceleration of the board will cause the stone to 
tip, supposing that the friction between the board and stone to 
be large enough to prevent slipping. 

246. Kinetic Reactions. — When a body rests upon a hori- 
zontal support, the reaction between the support and body 
equals the weight, but if the body is in motion, the reaction in 
general does not equal the weight. Thus, as seen in ex. 12, art. 
245, when an elevator moves with an acceleration a, the pres- 
sure which it exerts upon a body and its floor equals W±ma 
(W, m, and a denoting the weight, mass, and acceleration of the 
body respectively), the plus or minus sign being used according 
as the acceleration is up or down. The reactions of the cranks 
on a coupling-rod of a locomotive when it is at rest each equal 
one-half the weight of the rod, but when it is in motion they 
have a different value. The difference depends in part on the 
acceleration as is shown below. 

The component of a force acting on a body which depends 
upon its acceleration is said to be "due to acceleration" and 
"due to inertia," and such are often called inertia forces. Some- 
times it is desirable to determine the components of a reaction 
which are dependent and independent of acceleration. To dis- 
tinguish them we shall call the former and latter kinetic and 
static reactions respectively. No new principles are necessary 
for the computation of these. 



EXAMPLES. 

1. Suppose that the mass of the slider represented in fig. 187 
is m, the length of the crank c, and the number of revolutions 
per unit time n. Compute the crank-pin pressure at several 



24° 



TRANSLATION OF A RIGID BODY. 



[Chap. XL 



points of the stroke and plot curves showing how it varies with 
the displacement and the time. 

Solution : Neglecting friction the only horizontal force on the 
slider is the crank-pin pressure. Hence, denoting that pressure 
by Q and the acceleration of the slider by a, 

Q = ma, 



the positive direction being taken the same for Q and a. 

Since a= — 4x 2 n 2 csm 6= —^n 2 n 2 x (see eq. 3, art. 152), 

Q= —m^nWc sin 0= —m^n 2 n 2 x\ 



and the maximum value of Q is given by 
Q m =-m47t 2 n 2 c. 

Fig. 187 (b) and (c) shows graphically how Q varies with the dis- 
placement and the time. 




Fig. 187. 

2. Assume that the slider of the preceding example is verti- 
cal, that m = i8o lbs., c = io in., n = i20 rev.-per-min. Draw 
curves similar to those of fig. 187, showing the static and kinetic 
components of the entire pin reaction. 

3. Assume that the slider of fig. 187 is driven by steam pres- 
sure P, its value being known at each point of the stroke. As- 
suming that the crank turns uniformly, determine the crank- 
pin pressure for any position. 

Solution: (1) Algebraic. With notation as in ex. 1, 
P + Q=ma or Q=ma — P i 
a being equal to — 4?i 2 n 2 c sin or — 4n 2 m 2 x, as in ex. I. 



II.] 



APPLICATIONS. 



241 



(2) Graphical and employing D'Alembert's Principle. Since 
P, Q, and the reversed effective force for the slider are at .nail 
times in equilibrium (art. 238), Q is equal and opposite to the 
resultant of P and the reversed effective force. 

Let a'a" (fig. 188) represent the length of the stroke, and the 




Fig. 188. 



ordinates from a'a" to the curve c'c" the values of the steam 
pressure, P. From a' to a the pressure is forward and from a 
to a" backward. The effective force for the slider at each instant 
equals the product of its mass and acceleration ; it varies there- 
fore with the acceleration and the ordinates from a'a" to the 
straight line b'b" may represent the effective forces. During 
the first part of a stroke the acceleration is forward (in the direc- 
tion of the motion) and in the second part backward ; hence the 
reversed effective force in the first part of the stroke is backward 
and in the second forward. 

The directions of the steam pressure and reversed effective 
force from a' to b are opposite, from b to a the same, and from 
a to a" opposite again. Hence the resultants of the two forces 
from a' to b and from a to a" are represented by the differences 
between the ordinates to the two curves, and from b to a by their 
sum, i.e., by the ordinates of the shaded area. Obviously from 
a' to c the direction of the horizontal pin pressure is opposite tc 
the direction of the motion and from c to a" in that direction. 

4. Compute the kinetic reactions on a coupling or side rod 
of a locomotive running at a constant speed. 
Solution : Let the notation be as follows : 
r, radius of the crank-pin circles, v, speed of the locomotive, 
R, " " " drivers, m, mass of the rod. 

Since (by supposition) the acceleration of the locomotive is zero, 



242 TRANSLATION OF A RIGID BODY. [Chap. XL 

the absolute acceleration of the rod is the same as its accel- 
eration relative to the locomotive (see Prop., art. 196). Now 
the, motion of each point of the rod is just like that of the 
centre of a crank-pin, and relative to the locomotive that motion 
is circular, the speed being rv 2 /R 2 . When a point travels in a 
circle with a constant speed, its acceleration equals the speed 
squared divided by the radius of the circle and its direction is 
along the radius toward the centre (art. 193). Hence the accel- 
eration of the rod equals rv/R, and its direction at each instant 
is parallel to the crank. 

The effective force therefore equals mrv 2 /R 2 , and if the mass- 
centre is at the middle of the rod each kinetic reaction equals 
mrv 2 /2R 2 , its direction being from the corresponding crank-pin 
to the centre of the wheel. 

The total crank-pin pressure depends also on the weight of 
the rod and on the train resistance which is being "overcome." 

5. Assume that the weight of one side rod is 275 lbs., crank 
radius 1 ft., wheel diameter 5^ ft., and the speed 60 mi.-per-hr. 
Compute the crank-pin reactions due to acceleration and weight 
of the rod when it is at its lowest, highest, and middle posi- 
tions. 

247. Vibrations. — A study of vibrations furnishes applica- 
tions of equations of motion in which the force is variable. As 
illustrations, we choose vibrations caused by open coil-springs in 
various circumstances. If such a spring hangs vertically from 
one end and supports a body at the other, and the body is dis- 
placed vertically from its position of rest and then released, it 
will oscillate or vibrate up and down, the vibration enduring for 
a time and then ceasing. This is called a natural vibration, and 
in general a natural vibration is one executed by a body or sys- 
tem which has been displaced or distorted, then released and 
left to itself. 

If the support of the coil-spring is not fixed, but has a periodic 
up and down motion, the motions of the spring and suspended 
body are called forced vibrations. In general if a body or system 
executes a natural vibration and is then subjected to a periodic 
influence on its motion, the resulting vibration is called a forced 
one. 



§n.j 



APPLICATIONS. 



243 



The dying out of a vibration is ascribed to forces of the nature 
of friction ; thus in the preceding illustrations the forces are the 
resistance of the air upon the moving bodies and the internal or 
molecular friction in the spring. This effect (dying out) is 
known as damping, and vibrations in which damping occurs are 
called damped vibrations. For simplicity we will assume that 
there are springs without internal friction, and hence that such 
a spring and a suspended body might execute an undamped 
(or "simple") vibration. 

If an open coil-spring is elongated or compressed (not exces- 
sively), the elongating or compressing force is proportional to 
the elongation or compression of the spring. Strictly, this is true 
only when the act of elongation or compression is slow ; we will 
assume it to be true for a vibrating spring. No important error 
results if the mass of the suspended body is considerable as com - 
pared with that of the spring. Then if e and e' denote the elon- 
gations or compressions of a spring caused by forces T and T' 
respectively, 

T/T'=e/e' or T=(T r /e')e. 

248. Undamped Natural Vibration. — We use the following 
notation (see fig. 189): 

/ = natural length of the spring ; 
W = weight of the suspended body; 

m = mass " " " 

e' = elongation caused by W\ 

k = W/e'; 

y = displacement of the body from its position of rest ; 

a = acceleration " " at the displacement y ; 

T = force exerted upon the body by the spring at the 

displacement y. 
We choose the downward direction as positive for 
forces and displacements. 

The elongation (or contraction) of the spring for 
all positions of the body is (y-\-e f ) ; the spring is elon- 
gated or compressed according as iy + e') is positive 
or negative. Also 



I 

— * — -1 

IW I 

Fig. 189. 



T= -(W/e'){y + e f )=-(W/e)y-W t 



244 TRANSLATION OF A RIGID BODY. [Chap. XI. 

and the resultant force on the body in any position is 

W+T=-(W/e')y=-ky. . .... (i) 

Hence the equation of motion is 

— ky = ma, or a= —{k/m)y, .... (2) 

i.e., the acceleration is proportional to y (displacement) and is 
always directed toward a fixed point (from which y is measured) 
in the path; therefore the motion is simply harmonic (see art. 
179) and its period is 2iz/\/k/m. 

EXAMPLE. 

Suppose that W = 5 lbs. , e' = 3 in. , and that the body is released 
from a position 10 in. below that of rest. Describe the motion. 

249. Damped N atural Vibration. — The laws of the damping 
forces are various, depending upon the circumstances of the 
motion. If the vibrating system is like that represented in fig. 
189, the suspended body being a thin vertical plate immersed in a 
viscous liquid, then (neglecting the internal friction in the spring) 
the damping force consists of the fluid friction at the sides of 
the plate. If the velocity of the plate is small, this friction is ap- 
proximately proportional to the velocity. We choose this law 
of damping because the corresponding motion is analogous to an 
important electrical phenomenon. 

Let v and v' denote any two velocities of the immersed plate 
and F and F r the corresponding frictions; then 

F/F' = v/v' or F= (F'/v')v = cv, 

c being an abbreviation for F' '/v' '. If F be regarded as positive 
or negative according as it acts down or up on the plate, then 
because v = dy/dt (see the preceding article), F = cdy/dt, and 
the resulting force on the plate is 

W + T+F=-ky-cdyfdt. ..... (1) 

The equation of motion of the plate becomes, instead of eq. (2) 
art. 248, 

dy d 2 y 



- k y- c dt =m d t 



d 2 y c dy k 



(2) 



§ II.] APPLICATIONS. 245 

For brevity, let p = c/m and q = \/k/m\ then the solution of eq. 
(2) gives the following: 

(a) li\p*<q\ 

y = Ae~ ipt sin (Vq 2 -p 2 / 4 t + e) (3) 

e being the "Naperian base" and A and e constants of integra- 
tion depending upon "initial conditions." Thus let / = o and 
dy/dt = v when y = o, then substituting these in eq. (3) and in 
the expression for dy/dt, we find that 

£ = and A=v /\/q 2 —p 2 /4.. 

(b) Ifip 2 >q 2 , 

y = A£- at + B£-P\ (4) 

a and /? being abbreviations iov—p/2 ± \/p 2 / \—q 2 respectively 
and A and B constants of integration depending on initial con 
ditions. If as under (a), / = o and dy/dt = v when y = o, then 
substituting these in eq. (4) and in the expression for dy/dt, 
we find that 

A=v /(P-a) and B = v /(a-fi). 

EXAMPLE. 

Let ^ = 4 rad.-per-sec. and y = i6 ft.-per-sec. Plot on the 
same axes, 

(1) Equation (4), p being 9 rad.-per-sec. 

(2) Equation (3), p being 2 rad.-per-sec. 

(3) Equation (3), p being zero. 

'250. Undamped Forced Vibration. — Imagine the support of 
the spring in fig. 189 to oscillate up and down and call its dis- 
placement from the position shown x, regarding % as positive or 
negative, according as the displacement is down or up. The 
elongation of the spring at any instant is not y+ e' as in art. 248, 
but y + e' —x; hence 

T=-(W/e')(y + e'-x), 

and the resultant force on the suspended body is 

W + T=-k(y-x). ...... (1) 



246 TRANSLATION OF A RIGID BODY. [Chap. XL 

The equation of motion is 

— k(y — x)=m d 2 y/dt 2 ) 

or d 2 y/dt 2 + (k/m)(y-x)=o] (2) 

Now let the motion of the support be simply harmonic, its 

amplitude, period, and epoch being A, 2iz /to, and o respectively; 

then eq. (2) becomes 

d 2 y 

j+ + q 2 y = q 2 A sin a)t, ..... (3) 

q being an abbreviation for \/k/m, as in art. 249. The solution 
of the last equation is 

y= r-r^sinw/ (4) 

1 — (o 2 /q 2 

This shows that the forced vibration of the suspended body is 
simply harmonic, its period and epoch being the same as of the 
motion of the support, and its amplitude 1/(1— co 2 /q 2 ) times 
that of the support. Notice that co/q is the ratio of the fre- 
quency of the motion of the support to that of the natural vibra- 
tion of the spring. 

EXAMPLE. 

Let A = 1 , and plot a curve showing how the amplitude of 
the forced vibration varies for different values of aj/q, between 
1/10 and 10. 

251. Damped Forced Vibration. — Imagine that the suspended 
body of the preceding article is a vertical thin plate immersed in 
a viscous liquid. Then in addition to the forces acting on the 
body as described in the preceding article, there is the frictional 
resistance F= —cdy/dt (see art. 249). Hence the equation of 
motion is 

d 2 y c dy k dy k . . 
av m at m at m 



(1) 



p and q } being abbreviations as in art. 249. The solution of 

this equation is 

q 2 sin £ . , . v / v 

y= wp~ 4sin( ftrf-e), (2) 



I II. ] AP PLICA TIONS. 247 

£ being an abbreviation defined by 

tan £ = cop(q 2 — a> 2 ). 

This equation shows that the motion of the suspended body is 
simply harmonic, its period and amplitude being respectively 
the same as and (q 2 sin e)/ojp times that of the motion of the 
support and lagging behind the latter an amount equal to e. 

EXAMPLE. 

Assume that q = io rad.-per-sec. and that co varies from 7 
to 13 rad.-per-sec. Draw curves showing (a) how the lag e 
varies with co for values of p equal to 2, 1, 1/2, and 1/10, and 
(b) how the amplitude varies with cu for the four values of p 
just given. 

252. Kinetic Friction. — Definitions. The friction between 
two bodies which move relative to each other is called kinetic 
friction, and the ratio of the kinetic friction to the normal 
pressure between the bodies is called their coefficient of kinetic 
friction. If / denotes the coefficient and F and N the friction 
and normal pressure respectively, 

f = F/N or F = fN (1) 

Laws of Friction for Dry Surfaces. — (1) The coefficient de- 
pends on the nature of the rubbing surfaces. 

(2) The coefficient is approximately independent of the in- 
tensity of the normal pressure. Strictly it falls off slightly as 
the intensity increases up to a point when "seizing" is about to 
occur; then it increases rapidly. 

(3) The coefficient decreases as the velocity increases — not 
directly, but rapidly as the velocity increases from o to + 
and then less rapidly with increasing velocity. 

(4) The coefficient decreases with the lapse of time during 
the motion. 

These laws are qualitative; their reduction to a quantita- 
tive form has not been effected, but in some cases we know 
quite accurately how the coefficient varies, especially with 
velocity. Thus in certain experiments on the braking of rail- 



248 TRANSLATION OF A RIGID BODY. [Chap. XL 

way trains, the following values of the coefficient for the brake- 
shoe (cast iron) and the wheel (steel) were determined: 

Velocity in mi. -per-hr. . 17 21 27 31 37 47 
Coefficient 0.16 0.15 0.13 0.11 0.10 0.08 

These coefficients were taken five seconds after the brakes were 
set ; fifteen seconds after setting, the coefficients were about 
0.04 less than those given above. 

EXAMPLES. 

1. Fig. 190(a) represents a body resting on a plank, which in 
turn rests on rollers. If f is the coefficient of static friction for 




lseo. (b) 



Fig. 190. 

the body and plank, how large an acceleration can be given to 
the plank without causing slipping between it and the body ? 

Ans. f'g. 

2. Suppose that the body is slipping on the plank, and let 
f denote the coefficient of kinetic friction assuming it to be 
independent of velocity. How long (t) and how far (s) will the 
body slide in coming to rest from a velocity v under the influence 
of friction? Ans. t = v/f"g and s = v 2 /2f"g. 

3 . Suppose that the plank is caused to oscillate on the rollers 
by a force P, so that the acceleration-time curve for the motion 
is represented by fig. 190(6). Draw the velocity-time and space- 
time curves for the motion of the plank and also for the body. 
Where on the plank is the body at the end of one second ? 

Ans. It has slid forward on the plank 2^- ft. 



CHAPTER XII. 

ROTATION (RESUMED). 

| I. Second Moments of Mass (Moment of Inertia, etc.). 

253. Occurrence of Second Moments. — In a study of the rota- 
tion of a body, certain quantities are met with which are ex- 
pressed by integrals of the kind and form / dm • u 2 and / dm • uv, 

m denoting mass and u and v distances. Such quantities have 
been called " second moments of mass," the term being in line with 
"first moment of mass," which is applied to quantities expressed 

by integrals like / dm-x (see arts. 227 and 228). We distinguish 

between second moments of mass employing special names for 
the kinds. 

254. Moment of Inertia. — The moment of inertia of a body 
with respect to any axis is the sum of the products obtained by 
multiplying each elementary mass of the body by the square of 
its distance from the axis. The axis will often be called "inertia- 
axis" to distinguish it from other axes, coordinate, geometri- 
cal, etc. 

[Euler introduced the term "moment of inertia," and he explained its 
appropriateness somewhat as follows ("Thoria Motus Corporum Soli- 
dorum," p. 167): The choice of the name, moment of inertia (Ger. trdgheits- 
momenf) is based on analogies in the equations of motion for translations 
and rotations. In a translation, the acceleration is proportional directly 
to the "accelerating force" and inversely to the mass, or "inertia," of 
the moving body, and in a rotation the acceleration (angular) is propor- 
tional directly to the moment of the accelerating force and inversely to a 

quantity, I r 2 dm, depending on the mass, or inertia; this quantity, to 
complete a similarity, we may call moment of inertia. Then we have — - 

for translations, acceleration = (force) /(inertia, or mass) ; 

for rotations, acceleration = (moment of force) / (moment of inertia) J. 

249 



250 ROTATION. [Chap. XII. 

Expression for Moment of Inertia. — Let / * denote the moment 
of inertia of a body with respect to any axis, dm the mass of any 
elementary portion all points of which are equidistant from the 
axis, and p that distance. Then the definition states that 



fdmp' 



(1) 



the limits of integration being such that all the elementary parts 
of the body are included in the integration. 

If the body is homogeneous, d denoting its density and dV 
the volume of the elementary mass, dm = ddV\ hence 



l = dfdV 



P 2 (2) 



Units of Moment of Inertia. — From eq. (1) it is plain that 
the units involved in a moment of inertia are those of mass 
and length, and hence the unit of moment of inertia will de- 
pend upon the units of mass and length employed. No names 
are in use for the different units of moment of inertia, but each 
unit is described by stating the corresponding units of mass and 
length. Thus a moment of inertia computed by using the pound 
and foot is said to be expressed in a pound-foot unit. The unit 
corresponding to the geepound and foot may be called the engi- 
neers' unit of moment of inertia or the geepound-foot unit.f 

255. Radius of Gyration. — Since a moment of inertia is one 
dimension in mass and two in length, it can be expressed as the 
product of a mass and a length squared; it is sometimes con- 
venient to so express it. 

Definition. — The radius of gyration of a body with respect to 
an axis is such a length whose square multiplied by the mass of 
the body equals the moment of inertia of the body with respect 
to that axis. That is, if k and / denote the radius of gyration 
and moment of inertia of the body with respect to any axis and 
m its mass, 

k 2 m = I or k = VI7m. 

* A subscript affixed to the symbol refers to the inertia-axis ; thus l x 
stands for moment of inertia with respect to an x axis. 

t For dimensions of a unit moment of inertia, see Appendix C. 



I.J 



SECOND MOMENTS OF MASS. 



251 



The square of the radius of gyration of a homogeneous body 
with respect .to any axis is the mean of the squares of the dis- 
tances of all the equal elementary parts of the body from that 
axis. For let p lf p 2 , etc., be the distances from the elements, 
dm, to the axis, and let n denote their number (infinite). Then 
the mean of the squares is 

(p l 2 +p 2 2 + ■ ■ ■ )/n = (p 1 2 dm+p 2 2 dm-\- . . . )/n dm=I/m. 
But I/m is the square of the radius of gyration, hence, etc. 



EXAMPLES. 

1. Show that the moment of inertia and radius of gyration 
of a homogeneous right circular cylinder with respect to its geo- 
metric axis are respectively 

hnr 2 and rVi/2, 

m denoting its mass and r the radius of its base. 

Solution : Let a denote the altitude and imagine the cylinder 
to consist of elementary prisms parallel to the 
axis and extending from base to base. If dA 
denotes the area of the cross-section of any 
element, then dV = adA, and if p denotes the 
distance of the element from the axis, 

I = ojadA-p\ 

For convenience, select the prisms as indi- 
cated in fig. 191; then dA =pdp-dd, and 

/ = ad f Q r £ n p 4 dp • dd = ad^Ttr 4 = \mr 2 . 

Since k 2 =I/m, k = r\/i/2. 

2. Show that the moment of inertia and radius of gyration 
of a homogeneous parallelo piped with respect to one of its geomet- 
rical axes are respectively 




m(a 2 +b 2 )/ 12 and V{a 2 + b 2 )/i2, 

m being the mass and a and b the lengths of the edges which are 
perpendicular to the inertia-axis. 



; 5 2 



ROTATION. 



[Chap. XII. 



3. Show that the moment of inertia and radius of gyration 
of a homogeneous sphere with respect to a diameter are respec- 
tively 

2/$mr 2 and 7A/2/5, 

m being its mass and r its radius. 

Solution: Let fig. 192 represent a diametrical section of the 
sphere and Y the inertia-axis . Imagine 
the sphere to consist of elementary lam- 
inas perpendicular to OY ; then the mass 
of the lamina is given by 

dm = diz(r 2 — y 2 )dy. 

Now according to ex. 1, the moment of 

inertia of the lamina with respect to its' 

FlG - z 9 2 - geometrical axis (OY) is \dm{r 2 — y 2 ), 

or id7c(r 2 — y 2 ) 2 dy. Therefore the moment of inertia of all the 

laminas, or of the sphere, is the sum of all such moments, i.e., 

I = ^dnf^\r 2 -y 2 ) 2 dy 




A** 4 -* 



4. Show that the moment of inertia and radius of gyration of 
a homogeneous right circular cone with respect to its geometrical 
axis are ^respectively 

-f-^mr 2 and r\/ 3/10, 

m being its mass and r the radius of the base. 

5 . Show that the moment of inertia of a circular lamina with 
respect to a diameter of either base is approximately \mr 2 , 
m and r being its mass and the radius of the bases respectively. 

Solution : Let t denote the thickness of the lamina. Imagine 

it to consist of elementary prisms whose 

faces coincide with those of the lamina, 

their cross-sections being as shown in fig. 

193. The volume of each elementary 

prism is given by tpdd-dp. Now all parts 

of each elementary prism are not equally 

distant from the inertia-axis, but they are „ 

' J Fig. 193. 

nearly so, except in the case of prisms near 

the axis. These prisms, however, contribute little to the mo- 




§!•] 



SECOND MOMENTS OF MASS. 



253 



ment of inertia of the lamina and the error made in assuming the 
parts of such near prisms as equally distant from the axis is 
small. For any prism the distance is p sin d\ therefore approx- 
imately * 

/ = td£ *£sm dd ■ pHp = \nrHd. 

Hence, etc. 

6. Show that the radius of gyration of a thin elliptic plate 
with respect to either of its axes is \a, 2a being the length of the 
other axis. 

256. Relations Between Moments of Inertia and Between 
Radii of Gyration with Respect to Parallel Axes. — Proposition. — 
The moment of inertia (/) with respect to any axis equals its 
moment of inertia (7) with respect to a parallel central axis plus 
the product of the mass (m) and the square of the distance (d) 
between the axis, that is, 

I = I+md 2 



(1) 



Proof: Let fig. 194 represent 
a section of the body perpen- 
dicular to the inertia-axis. Let 
and C be the points where that 
axis and the parallel central axis 
respectively pierce the section. 
Then (see the figure) 




Fig. 194. 



I = jdm-p 2 , and since p 2 = y 2 + {x + d) 2 , 
I = fdm(y 2 +x 2 + 2xd + d 2 ) 
= I dm(x 2 +y 2 )+2d I dm-x + d 2 I dm. 
Now I dm{x 2 + y 2 ) = I ; / dm • x = mx = o ; and / dm = m ; hence , 



etc. 



or 



Corollary: Dividing the sides of eq. (1) by m we get 
I/m = l/m+d 2 , 
k 2 = k 2 + d 2 , 



(2) 



* The approximation is the closer the less the thickness. 



2 54 ROTATION, [Chap. XII. 

k denoting radius of gyration of a body with respect to any axis, 
k that with respect to a parallel central axis, and d the dis- 
tance between the axes. 

Equations (i) and (2) show that the moment of inertia and 
radius of gyration of a body with respect to a central axis are 
less than for any other parallel axis, and that approximately, 
when k is small compared to d,k = d and I = md 2 

EXAMPLES. 

1 . Determine the moment of inertia of a homogeneous par- 
allelopiped, the lengths of its edges being a, b, and c and its mass 
m, with respect to the third edge. Ans. m(a 2 +b 2 )/^. 

2 . Determine the radius of gyration of a homogeneous sphere 
with respect to a line whose distance from the centre is x. 

3 . Determine the radius of gyration of a right circular cylin- 
der with respect to a line parallel to its axis distant 10 ins. there, 
from, the radius of the base being 4 ins. 

4. Compute the radius of gyration of a rod whose cross-sec- 
tion is 1 X 1 in. and 24 ins. long with respect to a line parallel to 
a i-in. edge and 12 ins. from the centre of the rod. 

5 . Show that the moment of inertia of a right circular cylinder 
with respect to a central axis parallel to the bases is m(r 2 /^ + a 2 /T ) ) > 
m denoting mass, a altitude, and r radius of the base. 

Solution: Imagine the cylinder to consist of elementary cir- 
cular laminas. Call the distance of any one of them from the 
inertia- axis x, then its .thickness is dx and its volume is nr z dx. 
According to ex. 5, art. 255, the moment of inertia of a lamina 
with respect to its central axis which is parallel to the inertia- 
axis is \dmr 2 , and according to eq. (1), its moment of inertia 
with respect to the inertia-axis of the cylinder is 
\dm • r 2 + dm • x 2 = \dizr 2 dx + dnr 2 x 2 dx. 

Therefore the moment of inertia of all the laminas, or of the 
cylinder, is given by (if a denotes the altitude), 

I = 2d^(w£ /2 d X + f o a/ \Hx) 
= 2dTzr\^r 2 a + ^a 3 ); etc. 



I] 



SECOND MOMENTS OF MASS. 



255 



6. Show that the moment of inertia of a right elliptic cylinder 
with respect to a central axis parallel to either axis of the base is 
nt(a 2 /4+h 2 /s), 2a being the length of the other axis of the base 
and m and h the mass and altitude of the cylinder respectively. 

7. Show that the moment of inertia and radius of gyration of 
a hollow right circular cylinder with respect to its geometrical 
axis are respectively 

\m{r 2 + r 2 ) and V(r 1 2 + r 2 2 )/ 2 , 
m being its mass and r x and r 2 the inner and outer radii of its 
bases. 

257. Composite Bodies. — We refer now to bodies which can 
be divided into simple component parts; thus, a flywheel con- 
sists of hub, spokes, and rim whose forms are usually simple. 
The moment of inertia of such a body with respect to any axis 
can be computed by adding the moments of inertia of the com- 
ponent parts taken with respect to that same axis. 

EXAMPLE. 

Compute the moment of inertia and radius of gyration of the 
cast-iron flywheel (weight 450 lbs.-per-cu. ft.) represented in fig. 
195, the spokes being four in number and elliptic in cross-section. 



-10- 



^ 



iX'le— 



<-2- 




Fig. 195. 
258. Experimental Determination of Moment of Inertia. — 

There are a number of methods; we give two. 

Pendulum Methods. — (1) Suspend the body from an axis 
coinciding with or parallel to the inertia-axis. Let it oscillate 
like a pendulum noting the ' 'time " of an oscillation ( T) , and deter- 
mine the distance (a) from the axis of suspension to the centre of 
gravity of the body. Then substitute these values in the equation 

T = Wk 2 /ag* or k = (T/7z)\/gJa~ 

* This is the formula for the time of oscillation of a pendulum (see 
art. 267). 



256 ROTATION. [Chap. XII- 

(g denoting the acceleration due to gravity) , and solve for k. This 
is the radius of gyration of the body with respect to the axis of 
suspension, and from which the methods of art. 256 the 
desired moment of inertia can be computed. 

(2) From the same axis about which the suspended body 
swings, suspend by means of a cord a body whose dimensions 
are small compared with the length of the cord. Adjust the 
length of the cord so that the times of oscillation of the two sus- 
pended bodies are equal. Then measure the distance (I) from 
the axis of suspension to the centre of gravity of the small body 
and solve for k in the equation 

k 2 /a = l or k = VaT* 

k and a having the same meanings as in (1). 

Torsion-Balance Method. — There are many variations of the 
method here given. The balance for the present purpose may 
be arranged as follows: Suspend an elastic wire vertically, mak- 
ing the connection between the wire and support rigid , and fasten 
a flat plate in a horizontal position rigidly to the lower end of 
the wire — the form of the plate to be such that its moment of 
inertia with respect to the wire can be computed. 

Place the body whose moment of inertia is desired and a 
second body on the plate so that they " balance," i.e., leave the 
plate horizontal. The first body is to be so placed also that the 
wire is parallel to the axis with respect to the inertia-axis and 
the form of the second body is to be such that its moment of 
inertia with respect to the wire can be computed. Now, cause 
the loaded balance to oscillate and note the time (T) of one oscil- 
lation; then remove the two bodies and cause the empty balance 
to oscillate, noting the time (T ± ). 

Let I lt J 2 , and / denote the moments of inertia with respect 
to the wire of the plate, of the second body, and of the one 
whose moment of inertia is desired respectively; then, as shown 
in art. 268, 

r:T 1 ::\// 1 +/,+/:V37, 

or /=/ 1 ryr 1 2 -(/ 1 +/ 2 ). 

* Proved in art. 267. 



§11.] GENERAL PRINCIPLES. 257 

259. Product of Inertia. — Definition. — The product of inertia 
of a body with respect to two coordinate planes is the sum of 
the products obtained by multiplying each elementary mass of 
the body by its ordinates from those planes. 

Expression for Product of Inertia. — Let / denote the product 
of inertia of a body with respect to any two planes, as the yz 
and zx coordinate planes, dm the mass of any " third order " ele- 
mentary volume, and x, y, and z its three coordinates; then the 
definition states that 

J = /dm.xy, 

the limits of integration being so assigned that all the elementary 
parts of the body are represented in the integration. 

Since x and y have signs, the product dm-xy may be positive 
or negative, and hence a product of inertia may be (unlike a 
moment of inertia) negative or zero. 

260. Principal Axes. — It can be proved that through any 
point of a body there are three mutually rectangular axes, with 
respect to two of which the moments of inertia are greater and 
less than with respect to any other axis through the point. The 
three axes are called the principal axes of the body at that point. 

The condition that a line may be a principal axis of a body 

at some point of its length is that/ dm-xz and / dm-yz equal 

zero, the line being regarded as a z axis and the point as origin. 
(Proof must be omitted.) It follows from the foregoing that 

( 1 ) An axis of symmetry of a homogeneous body is a princi- 
pal axis at every point of it. 

(2) Any line perpendicular to a plane of symmetry of a homo- 
geneous body is a principal axis at the point where it pierces that 
plane. 

§11. General Principles. 

261. The Effective Forces. — Each particle of a rotating body 
revolves in a circle whose plane is perpendicular to the axis of 
rotation; therefore the acceleration of, and hence the effective 
force for, each particle has no component along the axis. 




2 5 8 ROTATION. [Chap. XII. 

Let the irregular outline (fig. 196) represent a rotating 
body, the axis of rotation being perpendicular to the plane 

of the figure at 0, and P any par- 
ticle; also let 
dm denote the mass of P ; 
r its distance from the axis ; 
a " acceleration; 

a the angular acceleratio of then 
Fig. 196. body; 

w its angular velocity. 
Then the effective force for P is dm -a, its direction being the 
same as that of a and the tangential and radial components of 
the force are respectively (since a t = ra and a n = rco 2 , art. 216) 

dm-m and dmroA 

262. Moment of the Effective System. — Of the two compo- 
nents of the effective force for any particle, only the tangential 
one has a moment about the axis ; hence the sum of the moments 
of the effective forces is 



/ (dm-ra)r = a I dm-r 2 = Ia, 



I being the moment of inertia of the body with respect to the 
axis of rotation. 

263. Equations of Motion. — According to D'Alembert's prin- 
ciple, the sums of the moments of the external and effective forces 
are equal. Hence if IM denotes the sum of the moments of the 
external forces about the axis for any instant, 

IM = Ia, M . ....... (1) 

a being the angular acceleration of the body at that instant, 
This is the equation of the motion of the body. 

As shown in art. 239 the sum of the components of the effec- 
tive forces along any line equals the product of the mass of the 
body and the component of the acceleration of the mass-centre 
along that line. We wish this sum for three lines, the tangent, 
the normal to the path of the mass-centre at the mass-cen- 
tre, and the axis. These sums are respectively 

mdt = m~a, ma n = m?a> 2 , and o, 



§ II.] GENERAL PRINCIPLES. 259 

wherein m denotes the mass of the body; 

a t the tangential acceleration of the mass-centre; 

a n '-' normal acceleration of the mass-centre; 

r " distance from the mass-centre to the axis. 
If IT, IN, and I A denote the algebraic sums of the compo- 
nents of the external forces along the tangent, normal and axis 
respectively, then according to D'Alembert's Principle, 

IT = mra, J'N^mrV, IA = o (2) 

These equations may be called the equations of motion of the 
mass-centre, since they involve terms depending on its motion. 
However, they are useful not to determine motion but especially 
to determine forces when the motion is known. 

264. Resultant of the Effective System. — It is assumed in this 
article that the rotating body is homogeneous and that it has a 
plane of symmetry perpendicular to the axis of rotation. Then 
the axis of rotation is a principal axis of the body where it pierces 
the Diane of symmetry (art. 260). 

Imagine the body divided into elementary rods parallel to 
the axis of rotation and then one of these rods into elementary 
portions of equal length. These portions have at any instant 
the same acceleration, and hence the effective forces for them are' 
equal and have the same direction. It follows that the resultant" 
of these effective forces is a single force whose action line is in the 
plane of symmetry. The effective forces for all the rods there- 
fore constitute a coplanar system, its plane being the plane of 
symmetry. The resultant of a coplanar system of forces is in 
general a force, and in special cases a couple (arts. 44 and 45); 
we proceed to determine these. 

Let fig. 197(a) represent the plane of symmetry of a body 
perpendicular to the axis of rotation, the intersection of the 
axis and the plane, and C the mass-centre. Also in addition to 
the preceding notation let 

k the radius of gyration of the body with respect to the axis ; 

a the acceleration of the mass-centre ; 

R denote the resultant effective force ; 

Ri its tangential component (JL to 0C)\ 

R n " normal component (|| to OC)\ 



260 



ROTATION. 



[Chap. XIL 






(a) 



According to art. 263 the sums of the components of the effec- 
tive forces along and perpendicular to OC equal respectively 
ma t = mra and ma n = mra> 2 . Therefore the components of the 
resultant effective force and the force itself are given by 

Rf = mra, R n = mro; 2 , R = ma, 

and the direction of R is the same as that of the acceleration of 
the mass-centre (a). 

The action line of the resultant cuts the line OC at a point Q 
whose distance from O is k 2 /r . This may be proved as follows : 
Imagine the resultant force to be resolved into its two compo- 
nents (mra and mrco 2 ) at Q. The moment of the force about O 
equals (mra)q. But the moment of the force equals the sum 
of the moments of its components, and this was shown in art. 
262 to be la; hence triraq = Ia, or 

q-k»/r. 

Three Special Cases. — (a) The angular velocity is constant, 
a = o. The resultant effective force equals nira> 2 , acts toward 
the axis and through the mass-centre (see fig. 1976). 

(b) The axis of rotation contains the mass-centre, r = o. The 
resultant force equals zero, but q =x , i.e., the resultant is a couple 
and its moment is la (see fig. 197c). 

(c) The axis contains the mass-centre and the angular veloc- 
ity is zero. The resultant vanishes completely.* 



* It can be shown that the resultant of the effective system is a force 
or a couple whenever the axis of rotation is a principal axis of the rotat- 



§ 111.] APPLICATIONS. 261 

265. Centripetal and Centrifugal Force. — When the resultant 
of the effective system is a single force, that of the external sys- 
tem is also a single force, and these resultants are identical; 
hence if C denotes the component of the latter resultant along 
the line joining the centres of rotation and mass, 

C=mfo> 2 =mv 2 /r. 

The component C of the external forces acting on the rotat- 
ing body is called centripetal force and the reaction correspond- 
ing to it is called centrifugal force. Observe that the first is 
exerted on the rotating body by other bodies and the second by 
the rotating body on the others, and that the centripetal force 
acts from the mass-centre towards the axis of rotation and the 
centrifugal in the opposite direction. 



§ III. Applications. 

266. Determination of the Motion. — If all the external forces 
which have a moment about the axis of rotation are given for 
any instant during the motion, the angular acceleration of the 
body at that instant can be computed from the equation of 
motion (art. 263). If the acceleration can be thus determined 
for each instant and the initial conditions of the motion (i.e., the 
angular velocity and the position of the body for any one instant) 
are also given, the motion can be completely determined. 



EXAMPLES. 

1. A body whose moment of inertia is I is made to rotate 
about an axis through the mass-centre by a constant force P 
applied to a cord wrapped about a cylindrical portion of the 

ing body at some point of the axis, and that, in this case, all the results 
of art. 264 hold if fig. 197 represents a section of the rotating body 
perpendicular to the axis of rotation, O the point where the axis is prin- 
cipal, and C the projection of the mass-centre on the section. 



262 ROTATION. [Chap. XII. 

body, as shown in fig. 198. Determine the motion, neglecting 
___^. axle friction and supposing that the angu- 

/~ ^n lar velocity and the position of the body at a 
// "V 1 certain instant are known. 
/ ( 0---A \ Solution: The external forces on the body 

\ V^_^/ J are P, its weight, and the reactions of the 
\^^~~ supports on the axle. Of these only P has 

a moment about the axis, hence the equa- 
tion of motion becomes 

Pr = Ia or a = Pr/I. . . . . . ( 1 ) 

This equation shows that the angular acceleration is constant, 
and for a given "turning moment" (Pr) applied to different 
bodies, their angular accelerations are inversely proportional to 
their moments of inertia with respect to the axes of rotation. 

Let the position of the body be specified by means of the 
angle d which a fixed line of it makes with a fixed reference line 
as in art. 209, and suppose that when P begins to act the angle d 
is zero and the angular velocity (cm) is also zero. Since a = dco/dt, 

dco = (Pr/I)dt or co = (Pr/I)t + C v 

If time is reckoned from the instant when P begins to act, &> = o, 
when t = o. Substituting these values in the last equation we get 

= + 6^, or C 1 = o J 
and hence w = (Pr/I)t. . . . . . . . (2) 

Since co = dd/dt, 

dd = (Pr/I)tdt, or d = $(Pr/I)t 2 + C 2 . 

Now d = o when t = o, and these values substituted in the last 
equation give 

= + C 2 , or C 2 = o, 

and hence d = %(Pr/I)t 2 (3) 

2. Suppose that the body in ex. 1 is a right circular cylinder 
of cast iron (weight 450 lbs.-per-cu.-ft.) 4 in. thick and 2 ft. in 
diameter, and that P is applied at the rim, its value being 6 lbs. 
Determine the acceleration. Ans. 1.64 rad./sec. 2 

3. Suppose that at a certain instant the wheel of ex. 2 is 
rotating at 20 rev.-per-sec. in the counter-clockwise direction. 



§ HI.] APPLICATIONS. 263 

What is its velocity 10 sees, later, P acting upon the wheel dur- 
ing that time? 

4. Solve ex. 1, supposing that a body whose weight is W is 
suspended from the cord wrapped around the drum. 

Solution: Let T represent the tension in the cord. This is 
the force replacing P of ex. 1, and the equation of motion for the 
rotating body is 

Tr = Ia (1) 

This contains two unknowns T and a; to determine them we 
write next the equation of motion for the suspended body. The 
external forces on it are T and W, and as its acceleration is 
downward the resultant force on it is down, i.e., W is larger 
than T; hence 

W—T = ma, (2) 

m and. a being the mass and acceleration of the suspended body. 
These two equations contain three unknowns, T, a, and a, but 
we have the following additional relation : 

a = ra (3) 

From these three equations we find that 

r r 2 

and 



r 2 + k 2 * r 2 + k 2 *' 

k denoting the radius of gyration of the rotating body with 
respect to the axis of rotation. The equations also determine 
the value of the tension ; thus 

T = W-mr 2 g/(r 2 +k 2 ). 

5. Suppose that a wheel rotates about a horizontal axis "out 
of centre" as represented in fig. 199, and that the only external 
forces on the wheel are its own weight and 
the reaction of the supports (no friction). /^^ ^^\ + \ 
The angular velocity when C is directly to / ,'\ 

the right of being given as co Q counter- [ ~^ \$ | 

clockwise, determine the angular velocity of y ! — j — 

the wheel in any position. \. v y 

Solution : The equation of motion is -—^'^ 

-W~r cos d = Ia=Id 2 6/dt 2 , FlG " I99 ' 



264 ROTATION. [Chap. XII. 

and integration of it gives 

-Wrsind = hIco 2 -hIcD \ 

from which oj can be computed for any value of d. 

6. Solve ex. 14 art. 245 taking into account the mass of 
the pulley (then the tensions on opposite sides of the pulley are 
unequal) , its radius of gyration with respect to the axis of rota- 
tion being k, its radius r, and its weight W. 

Ans a= (Wt-wte , 
• wk 2 +(w t +w 2 y 

7. Solve ex. 15 art. 245 taking into account the mass of 
the pulley, its radius of gyration with respect to the axis being k, 
its radius r, and its weight W. 

8. Solve ex. 16 art. 245 supposing that £ = ^7/9 ft., r=ij 
ft., W=i44 lbs., and £ = 32. 

9. Fig. 200 represents a tub floating upside down. Two 
cords are wrapped about the tub in opposite directions and lead 
off in parallel directions over pulleys as shown, and sustain 
bodies W and W. Discuss the motion of the tub under the 
influence of the suspended bodies and the fluid frictional resist- 
ance which assume to be proportional to the velocity. 




Fig. 200. 

Solution : Let 6 denote the angular distance described by the 
tub in any time t after starting, co the angular velocity, and a 
the angular acceleration. Let F denote the frictional resistance 
at any instant; then since it varies as the velocity, F^cco, c 
being a constant depending on the liquid, diameter of tub and 
extent of the wetted surface. Let T denote the tensions in the 
cords, evidently the same at any instant, but not constant in 
time. Also let I denote the moment of inertia of the tub with 
respect to the axis of rotation, 2T its diameter, m and W the 
mass and weight of the suspended bodies, and a their accelera- 
tion. 



*ni.] 



APPLICATIONS. 



265 



The equations of motion for each suspended body and the 
tub are 

W— T = ma and 2Tr — 2Fr = Ia. 

Combining these with a = ra, we get 

(/ + 2mr 2 )a + 2rca> = 2 Wr, 
and abbreviating, we have as the equation of motion 





A w +Ba " 


The first integration 


of this gives 




•-g(i-«- 


and the next 






'-b{' + f 



c. 



A" 
B. 



(D 



<») 



(3) 



10. Plot a curve showing how the angular velocity changes, 
taking A, B, and C as 2, 5, and 10 respectively. 

11. Suppose that there are no cords and suspended weights 
and that the tub is given an angular velocity w . Discuss the 
motion of the tub under the influence of fluid friction. 

Ans. oj = aj e~ bt (b being equal to 2rc/I). 
267. Pendulums. — A body which rotates about a horizontal 
axis under the influence of its 
weight and the reaction of the 
support is called a compound 
or physical pendulum. Let fig. 
201 represent a section of such a 
pendulum perpendicular to the 
"axis of suspension" and through 
the mass-centre C. Let O be the 
intersection of the axis of suspen- 
sion and the section, and let 
a denote the distance OC ; 
k the radius of gyration with re- 
spect to the axis of sus- 
pension; 
T '* time of one oscillation (from one extreme position to the 
other) ; 




266 ROTATION. [Chap. XII. 

d the angle XOC; 
6 ± " maximum value of d\ 
W " weight of the pendulum; 
m its mass. 

We regard the counter-clockwise direction as positive and as- 
sume that the support is frictionless, or has no moment about 
the axis of suspension. Then the equation of motion becomes 

— Wa sin d=mk 2 a =mte d 2 ti/dt 2 * 

or d 2 d/dt 2 =-(ag/k 2 )smd (i) 

The complete integration of eq. (i) is expressible by an infinite 
series, but it is not here given because we wish the value in a 
special case which admits of a simple approximate integration. 

We will assume that the amplitude of the oscillations (0 t ) is 
so small that practically sin d = 6 ; then eq. ( i ) becomes 

d 2 d/dt 2 =-(ag/k 2 )d, (2) 

the first integration of which gives 

(dd/dt) 2 = -(ag/k 2 )d 2 + C X . 

Now when 6 = d lt dd/dt = o; therefore substituting these values 
in the last equation we find that 

o=-(ag/k 2 )d 1 2 + C 1 , or C x = {aglh 2 )d 2 , 
and (dd/dt)=±(ag/k 2 )\d l 2 -d 2 )l (3) 

The plus or minus sign is to be used according as dd/dt is posi- 
tive or negative, i.e. , according as the pendulum is swinging in the 
positive or negative direction. The integration of eq. (3) gives 

sin" 1 (0/#i)= ±(ag/k 2 )h + C 2 . 

Now if we reckon time from the instant when the pendulum 
passes through its lowest position, i.e., t = o when d = o, the last 
equation becomes for these values 

sin -1 (o) = o + C 2 ; hence C 2 = o. 
If when * = o the pendulum is moving in the positive direction, 

sin- 1 (^ 1 ) = (a^ 2 )^ ) 

(4) 



or d = d x sm(Vag/k 2 t) 



§ III.] 



APPLICATIONS. 



267 



This equation is analogous to that for a simple harmonic motion 
(see art. 179), and hence the motion of a pendulum is often 
called "a simple harmonic oscillation." 

The time of an oscillation can be found from eq. (4). Let 
t x and t 2 denote the values of t when the pendulum is in its high- 
est positions on the right and left respectively. Then when 
t=t ti d = d Xi and when t = t 2 , it can be shown that d= — d lt hence 



t x = Vk 2 /ag 



sin 



and 



= iWk 2 /ag, 



t 2 = Vk 2 /ag sm-\ 

or T = -Vk*/ag* , (5) 

This expression for T being independent of d, shows that the 
time of oscillation of a pendulum is the same for all values of d 1} 
provided that it is so small that sin d 1 practically equals V 

The point Q (in OC) whose distance from equals k 2 /a 
is called the centre of oscillation and the line through it parallel 
to the axis of suspension is called the axis of oscillation. The 
following is a simple geometrical construction for locating the 
centre of oscillation: Let O (fig. 202) be the centre 
of suspension and C the mass-centre; then OC = a. 
Let k denote the radius of gyration of the pendu- 
lum with respect to a central axis parallel to the 
axis of suspension and lay off CK equal to k; join 
O and K and draw KQ perpendicular to OK; then 
Q is the centre of oscillation. For 

CK 2 = OC-CO, or CQ = k 2 /a, 
hence OQ = a + k*/a = {a 2 + k 2 )/a = k 2 /a 
(see eq. (2), art. 256). 

* Experiment Proof that the Masses of Bodies are Proportional to their 
Weights at the Same Place. — The expression for the time of oscillation of 
a pendulum was deduced on the assumption that mass is proportional to 
weight, for we substituted g for W/m in deducing eq. 1. If this substi- 
tution is not made eq. (5) becomes T= -k\/m/aW. 

Now take two pendulums, each consisting of a sphere suspended by 
means of a light cord, the lengths of the cords being the same; also, to 
make the air resistances the same, the spheres should be equal in size. 
Next, compare their times of oscillation; it will be found that they are 
equal. Call this time T, and the weights and masses of the pendulums 




268 ROTATION. [Chap. XII. 

By length of a compound pendulum for any given axis of sus- 
pension is meant the distance from that axis to the centre of 
oscillation. It is shown in the next paragraph that this is the 
length of an equivalent "simple pendulum." 

A simple or mathematical pendulum is an ideal one consisting 
of a particle suspended by means of a massless cord. Evidently 
this is a special form of the physical pendulum, and the preced- 
ing discussion applies. Let / equal the length of the cord, then 
for the simple pendulum 

k = a = l, or k 2 /a = l, 
hence T = 7i\/Tfg (6) 

EXAMPLES. 

i. A "seconds pendulum" (one whose time of oscillation is 
one second) is found to be 39.12 in. long at a certain place. 
What is the value of g at that place ? 

2. Show that the "axes of suspension and oscillation are in- 
terchangeable," i.e., that the times of oscillation are the same 
whether a pendulum oscillates about or about Q. 

3. Show that on any line OCQ there are four points or axes 
about which the pendulum will oscillate with the same time T. 

4. According to the law of gravitation (art. 87) the attrac- 
tion of the earth and hence the acceleration due to it varies in- 
versely as the square of the distance from the earth's centre. 
Or if g t and g 2 denote the accelerations at two points whose dis- 
tances from the centre are r and r + e respectively, 

gi/g^(.r+ey/r\ 
Show that if T t and T 2 are the times of oscillation of a pendulum 
at the two places respectively, approximately 

T 1 /T 2 = i-e/r, and e = r(i-TJT 2 ). 

W 1 and W 2 , and m l and m 2 , respectively. Evidently k has the same 
value for the two pendulums; also a. Hence the equation above be- 
comes for the two pendulums 

T=itk Vm 1 /aW 1 and T = nk Vm 2 /aW 2 
or m 1 /m 2 = W l /W 2 . 

This is practically the method employed by Newton; he used hollow 
wooden spheres containing gold, silver, lead, glass, sand, common salt, 
wood, water, and wheat. 



I HI.] APPLICATIONS. 269 

268. Torsion Balance. — When a torsion balance (see art. 258) 
is displaced through a small angle (the "pan" being rotated 
about the wire), the moment of the couple required to produce 
the displacement is proportional to the displacement. Thus if 
M and M' denote two values of the displacing couple and 6 
and d' the corresponding displacements 

M/M' = d/d', or M = Cd, 

C being an abbreviation for M'/Q'. 

The wire exerts upon the pan in any displaced position a 
couple whose moment is equal but opposite in sign to that of 
the displacing couple. 

Imagine the pan displaced an amount U and then released; 
it will oscillate under the influence of the couple which the wire 
exerts upon it. If I denote the moment of inertia of the pan 
(and its contents if any) with respect to the axis of the wire, 
then the equation of the motion is 

-C6 = Ia, or d 2 d/dt 2 = (-C/I)6 (1) 

This is analogous to eq. (1) art. 267, therefore its soli 1 1 ion is 
left to the student. He should find that the time of oscillation 
(7) is given by 

T = WIJC (2) 

269. Conical Pendulum. — A conical pendulum consists of a 
body suspended from a fixed point by a cord and so that it can 
be made to rotate about the vertical axis 

through the fixed point (see fig. 203). The 
motion might be caused and maintained by I 

means of a vertical board rotating about / 

the axis and pressing laterally against the if 
suspended body. We wish to determine / 
the relation between the angular velocity /^_ 
(w) of the body when constant and the \~ 
44 height" (h) of the pendulum. "" 

Let P denote the tension in the string; Fig. 203. 

W the weight of the body ; 

m its mass ; 

R the pressure of the board against the body. 



270 



ROTATION. 



[Chap. XII 



Then neglecting air resistance, eqs. (2) art. 263, become 
IT = R = o. .......... 

IN = P sin <j> = mra> 2 =ml sin <f>-oj 2 . . , 
IA=P cos <j>-W = o. . . . . . 

From (2) and (3) we find that h = g/w 2 . 



CD 

(3) 



EXAMPLE. 

Suppose that the weight of the rotating body (fig. 203) is 
10 lbs., that it makes 100 rev.-per-min. and / is 15 in. What 
are the values of h and P? 

270. Weighted Conical Pendulum Governor. — This consists 
of three heavy bodies, A, B y and C (fig. 204), connected by light 

links as shown, the whole system 
being supported at D and revolv- 
ing about a vertical axis AD. We 
wish to determine the height (h) 
for a given angular velocity (co). 
Let Pj denote the force exerted on 
B by BD\ 
P 3 the force exerted on Bby BA : 
W " weights of B and C; 
W ± " weight of A. 
Then the forces exerted upon A 
and B are as shown in fig. 204(6). 
Eqs. (2) art. 263 for A become 

IA = 2P 2 cos<t>-W l ^o i (1) 
and for B, 

i^iV = P 1 sin^ + P 2 sin0 = (W/g)/sin^-^, . . (2) 
1A=P ± cos <f>-P 2 cos cf>-W = o. ..... . (3) 

Solution of these three equations gives 

2{W + W l )g , 

n- Wa>2 . ......(,. 





(6) 






Fig. 204. 



EXAMPLE. 

Let W t = 10 and W = 8 lbs., and draw a curve showing how h 
varies with the number of rev.-per-min 



ELEVATION 



R., 



§IIL/j APPLICATIONS. 271 

271. Kinetic Reactions. — Definition {repeated, from art. 246). 
— By the kinetic reactions upon any body is meant such com- 
ponents of the forces acting upon it which depend upon its 
acceleration. The determination of kinetic reactions of rotat- 
ing bodies is illustrated in the solution of some of the follow- 
ing 

EXAMPLES. . 

1. A cubical box, into which a sphere just fits without pres- 
sure, is made to rotate about a ver- 
tical axis, as shown in fig. 205. 
Determine the kinetic reactions on 
the sphere when the angular ve- 
locity and acceleration are a> and 
a respectively. 

Solution: Let R t denote the 
pressure of the bottom of the box, l* 3 

R 2 that of the outer side, and R 3 FlG - 2 °5- 

the third one. Evidently the latter acts as shown if the accel- 
eration is counter-clockwise. Equations (2) art. 263 become, 
if W and m denote the weight and mass of the sphere respec- 
tively, and v the velocity of the mass-centre, 

JT = R 3 = mra, 

UN = R 2 = mrco 2 = mv 2 /?, 

ZA=R 1 -W = o, or R X = W. 

These show that R 2 and R 3 are entirely kinetic and that R t is 
static. 

2. In ex. 1, let W=io lbs. and suppose that the angular 
velocity increases every minute by 2 rev.-per-min. Determine 
the kinetic reactions when the angular velocity is 10 rev.-per- 
min. 

3. A body rests upon the floor of a car which moves in a 
horizontal circular curve of radius r with a constant speed v. 
Determine (1) the kinetic reaction on the body and (2) the 
direction of the resultant pressure of the body on the floor. 

Ans. (2) Inclination to the vertical, taxr l (y 2 /rg). 




272 ROTATION. [Chap. XIL 

4. A body is suspended by means of a cord from the ceiling 
of a car which moves in a horizontal circular curve of radius r 
with a constant speed v. Determine the direction of the sus- 
taining cord and the tension in it. 

5. Fig. 206 represents a car on a tilted track. Suppose that 
the track is a horizontal circular curve of radius r and that the 

car moves with a constant speed v. Deter- 
mine the kinetic reaction on the car and 
the angle of tilt which makes the resul- 
tant of the flange pressures * zero. 

Solution : Imagine each wheel pressure 
resolved into three components, one par- 
allel to the rails, one parallel to the ties, 
and one perpendicular to the first two. 
~T. ~~ Call the sums of these components R', 

R" ', and R ,n respectively, and the resul- 
tant of R" and R"' R\ also let P' and P" denote the pulls at 
the front and rear of the car respectively, which assume to be 
practically parallel. 

Since the velocity of the car is constant, a = o and (see eqs. 
(2) art. 263), 

IT = P'-P"-R' = o, 

2N = R" cos <f>+R'" sin <t> = mv 2 /r, 

2 A =R"' cos tj>-R" sin cj>-W = o. 

These equations show that R' = P' —P" 

and that R = (m 2 v 4 /r 2 + W 2 )±. 

If R" (the sum of the flange pressures) equals zero 

IN = R" sin <j> = mv 2 /r and I A = R" cos <j> - W = o ; 
hence, combining tan <f> = v 2 /gr.'\ 



* By flange pressure is here meant the component of the pressure oh 
a wheel parallel to a tie of the track. 

f This relation makes the sum of the flange pressures, but not each one 
necessarily, equal to zero. It has been discovered experimentally that if 
the wheels are coupled together in fours as usual, the front outer wheel 
always experiences a flange pressure. 



§ III.] 



APPLICATIONS. 



273 



These results can be reached graphically a little more simply 
as follows: Since R l ', P' , and P" are in equilibrium, R and W are 
equivalent to the effective force for the car, i.e., the resultant 
of R and W is identical with the resultant effective force. So 
draw from C a line,CV to represent the resultant effective force 
mrco 2 = mv 2 / V, a line Cw to represent W, and complete the par- 
allelogram Cwcr. Then Cr represents R, and from the figure it 
is seen that R equals the value given in the foregoing. In order 
that R may have no component along the tie, i.e., R" = o, 
the angle Crc must equal <f>, or 

tan <f> = Cc/cr = (mv 2 /r)/W. 

272. Weight of a Body as Influenced by the Earth's Rotation. — 
Let fig. 207 represent a meridional section of the earth, ON being 
the polar axis. Imagine a body resting on 
the surface at A or suspended by means 
of a cord. The forces acting upon the 
body are two in number, the attraction of 
the earth (P), and the reaction of the sup- 
port or the pull of the cord (Q). P is 
directed somewhat as shown; let its mag- 
nitude be represented by AB. Q is not 
collinear with P (except at the equator or 
pole) because the resultant of P and Q 
must be directed the same as the resultant 
effective force for the body; the direction 
of this is AD (the radius of the path of A). 

Let R denote the resultant effective force (also the resultant 
of P and Q), m the mass of the body, co the angular velocity of 
the earth, and r the radius of the path of .4 ; then 

R = mra> 2 , 

and if AD represents R, the side AC of the parallelogram drawn 
on ABD represents Q. It follows from the figure that 




Fig. 207. 



P* = Q*+R* + 2 QR cos <f>, 



or, 



Q = PVi -sin 2 (j)(R/P) 2 -R cos 0. 



• (1) 



i 74 ROTATION. , [Chap. XII. 

It is shown in ex. i below that R/P is less than 1/289; hence, 
approximately 

Q = P—R cos <j) = P—mrojl cos <f> (2) 

Q or its opposite is the force which we actually measure by 
spring- or beam-balance and call the weight of the body. It 
may be called "apparent weight" to distinguish it from the 
attraction P, or "real weight." Eq. (2) shows that the apparent 
weight is always less than the real, and that the difference de- 
pends on r cos <j>. 

Notice that <p is the latitude at A , since it is the angle made 
by the plumb line at A with the equatorial plane. Since R 
(=mrco 2 ) is equal to the centrifugal force of the body, the rela- 
tionship in eq. (2) is sometimes expressed thus: "The weight 
of a body is diminished by the product of its centrifugal force 
and the cosine of the latitude." 

EXAMPLES. 

1. Show that at the equator the difference between the real 
and apparent weights of a body is about 1/289 of the apparent 
weight. 

Solution: From eq. (1), since sin <^ = o at the equator 

(P-Q)/Q=R/Q = nir aj 2 /Q, 

r denoting the equatorial radius. Now m/Q = i/g, g denoting 
the acceleration due to gravity at the equator as measured ex- 
perimentally, i.e., g is also "apparent"; hence 

(P-Q)/Q = r aj*/g. 

Now oj = 27t/t, where / denotes the time of one revolution of the 
earth; and since 2 = 86, 164 sec, r = 20,920,000 ft., and £ = 32. 09 
ft. /sec. 2 , r co 2 /g = 0.003467 = 1/289. 

2. Show that if the earth rotated 17 times as fast as it does> 
then the apparent weight of a body at the equator would be 
practically zero. 

273. Centrifugal Hoop Tension. — Imagine a hoop to lie upon 
a horizontal table which rotates about a vertical axis through 
the centre of the hoop. The tension which exists at each cross 
section of the hoop is called "centrifugal hoop tension"; we 
now deduce an expression for it. 



§ in.] 



APPLICATIONS. 



275 




Let fig. 208 represent one half of the hoop. The forces act- 
ing on this half consist of its weight, the reac- 
tion of the table, and the forces exerted by the 
other half, i.e., the hoop tensions at the two 
sections. Since the first two forces balance 
each other, the resultant of the remaining two 
must equal the resultant effective force. Hence 
if m denotes the mass of one half of the hoop, 
r the distance from the axis to its mass-centre, 
a> its angular velocity, and P the hoop tension, Fig. 208. 

2P = mra> 2 , or P = \mrco 2 . 

EXAMPLE. 

Let r denote the radius of the hoop and w its specific weight. 
Regard the tension at a section as uniform (practically true 
when the thickness is small compared to the radius), and show 
that the intensity of the hoop tension equals 2wr 2 co 2 /g. 

274. Hinge Reactions. — Rotating bodies often turn (a) about 
a fixed shaft or (b) with a shaft in fixed bearings. The force 
exerted by the body on the shaft in the first case and those 
exerted by the bearing on the shaft in the second will be called 
hinge reactions. Determination of hinge reactions in the fol- 
lowing is limited to cases in which the rotating body has a plane 
of symmetry perpendicular to the axis. Then the resultant of 
the effective system for the body consists of a single force (see 
art. 264). 

Case I. Rotation about a Fixed Shaft. — We assume that the 
applied forces are such that the hinge reaction is equivalent to 
a single force which call R. Let fig. 209 represent a section of 

the rotating body through the 
mass-centre (C) and perpendic- 
ular to the axis (O). Imagine 
R resolved into three compo- 
nents, one parallel to OC* one 
parallel to the axis, and one per- 
pendicular to OC and the axis, 
and denote them by R n , R a (not shown), and R t respectively. 

* If the mass-centre is in the axis the direction of OC may be taken 
any way perpendicular to the axis. 




276 



ROTATION. 



[Chap. XII. 



Let 2F ny IF a , and !F t denote the algebraic sums of the com- 
ponents of all applied forces {i.e., all the external forces except the 
hinge reaction), parallel to R n , R a , and R t respectively. The 
components of the resultant effective forces parallel to these 
same directions are mrcu 2 , o, and mra, acting as shown in fig. 209. 
Since the external forces and the reversed effective forces are 
in equilibrium (art. 238), 

R t + IF t = mra , or R t = mra — IF ti 
R n + 2F n = mrco 2 , or R n = nvrco 2 — 2F n , 
Ra + ZF a = o* or R a =-IF a . 
These equations show that R a has no kinetic component, and 
that the kinetic components of R t and R n equal zero if 7 = o, 
i.e:, if the mass-centre is in the axis and if, as was assumed at 
the outset, the body has a plane of symmetry perpendicular 
to the axis. 

Case II. The Body Rotates about a Shaft in Bearings. — We 

assume that there are two bear- 
ings whose reactions call R' and 
R". Let A and B (fig. 210) be 
the bearings, and the parallelo- 
gram the plane of symmetry of 
the body. 

Imagine R' and R" (like R, 
Case I) resolved into three com- 
ponents, and extend the nota- 
tion of that case to the present 
one. Also let IM t and* ZM n 
denote the moment sums of the 
applied forces {all the external 
forces not including R' and R") with respect to the lines marked 
Oi and On respectively. Then as all the external forces and 
the reversed resultant effective force are in equilibrium, 

R/+R t " + lFt=nira, 

Rn'+Rn" + 2F n = ynrio 2 , 
Ra'+R a " + IF a = o, 

Rn"l"-Rn'l' + 2M t =0, 

R/l'-R t "l" + IM n = o. 




* These follow also from eqs. (2), art. 263. 



§ m.] 



APPLICATIONS. 



277 



From these equations we find that 

R t 'l = {mra - IF t )l" - IM n , . 
R n 'l = (mrco 2 - IF n )l" + 2M U 
R»l = (tnra - 2F t )V + IM n , 
R n "l = {mrco 2 - IF n )V - IM U 

R a '+R a "=-2F a - 

These equations show that the kinetic components of the hinge 
reactions are zero if r=o, i.e., if the mass-centre is in the axis, 
and if, as was assumed at the outset, the rotating body has a 
plane of symmetry perpendicular to the axis of rotation. An 
axis of a body for which the kinetic components of the hinge 
reactions are zero is called a "free axis." It can be shown that 
the three central principal axis of any body are free axes. 

EXAMPLES.* 

1. Suppose that in ex. 5 art. 266, W= 100 lbs., 1 = 2 (gee- 
pound-foot units), r = i/2 ft., and co Q = 4 rad.-per-sec. Compute 
the hinge reaction when C is directly to the right of O. 

Solution: According to the solution of ex. 
5, the angular acceleration in the position un- 
der consideration is — 5 rad.-per-sec. -per-sec. 
Hence (see fig. 211) 

R t -ioo= —(100/32. 2)-J«5, or R t = g2.2 lbs. 
-R n = (100/32.2) •£• 16 = 24.8 lbs. 

2. Determine the hinge reactions in ex. 1, 
vvhen C is vertically above O, below 0, and to the left of O. 

3 . Suppose that the wheel of the preceding example revolves 
about a vertical axis with a constant angular velocity of 4 rad.- 
per-sec. Determine the hinge reaction in any portion of the 
wheel. 

* The student is advised not to use the foregoing formulas, but to 
proceed as follows: (1) Determine the resultant effective force for the 
rotating body in the position under consideration, remembering that the 
normal component acts from the mass-centre toward the axis and the 
tangential component in the direction of the tangential component of 
the acceleration of the mass-centre. (2) Write the conditions of equi- 
librium for the force system consisting of that resultant reversed and aH 
the external forces (including the hinge reactions) . (3) Solve. 




Fig. 211. 



278 



ROTATION. 



[Chap. XII. 



4. Determine the hinge reaction on the rotating body de- 
scribed in ex. 2, art. 266. 

5. Imagine the rotating body of fig. 210 to be a parallelo- 

piped whose weight is 90 lbs., and 
whose radius of gyration with 
respect to a central axis parallel 
to the axis of rotation is 7.84 in., 
that /' = /" = 18 in., and OC=i in. 



C 



■0/ * 



15' 



■¥— 



o 

48 lbs. 



ts? 



m* 



90 lbs. 



o 

48 lbs. 



18' 



Fig. 212. 



dU ^c, I^^ Qj Suppose that the body is rotated 
by means of a couple applied to a 
thin disk (mass negligible), as 
shown in fig. 212. Determine the 
hinge reactions when the motion is 
about to begin (co = o). 

Solution: The masses of the 
suspended and rotating bodies are 
respectively 

48/32.2 = 1.49 and 90/32.2 = 2.79 geepounds. 

The moment of inertia of the rotating body about the axis of 
rotation is 

2.79 (7.84/i2) 2 + 2.79 (i/i2) 2 =i.2i (geepound-foot units). 

Hence the equations of motion of the suspended and rotating 
bodies are respectively, T denoting tension in the cord, 

48 — 7^=1.490, T-i = i.2ia; 

and since here a = a/2, we find from the equations that a = 24.5 
rad.-per-sec.-per-sec. 

The components of the resultant effective force are 

nira = 2.79 XyVX 24.5 = 5.68 and o, 

and act as shown in the figure. Since the reversed resultant 
effective force and the external forces are in equilibrium, we 
next write as many conditions of equilibrium for that system 
as are necessary to determine the unknowns. Thus, for the 
action line of R/' as moment axis, 



2?»'-3— 90-1/12 



or R n ' = 2.$\bs. 



§ in.] 



APPLICATIONS. 



79 



for the action line of R n " as moment axis, 

R/- 3-5.68-1^ = or ^' = 2.84 lbs.; 
and since R t '+Rt" = $.6$ and R n '+R n " = o, 

£," = 2.84 and R n " = -2.5 lbs. 
From the " axis resolution equation" 

R a "-go = o, or R a " = 90 lbs. 

6. Determine the hinge reactions when the body has rotated 
through 90 , 180 , 270 , and 360 , and record your results in a 
tabular form as follows: 



Re 



o 
90 c 



a 


<y 


mra 


mraj 2 


#/ 


#»' 


W 


#»" 


1 . 21 

1 . 21 





5.68 





2.84 


2 -5 


2.84 


- 2 -5 



90 



7. Suppose that in ex. 5 there is no couple applied and that 
the body rotates with a constant speed of 100 rev.-per-min. 
Determine the hinge reactions. 

8. Suppose that in ex. 7 there is a second parallelopiped 
just like the first attached to the shaft, also 1 in. "out of cen- 
tre," and so that the two mass-centres are on opposite sides of 
the shaft, the second one being 3 in. vertically above the first. 
When they are rotating at 100 rev.-per-min., determine the 
hinge reactions. 

275. Balancing of Rotating Bodies. — As illustrated by exs. 
5-8 of the preceding article, when one or more bodies rotate 
with a shaft, the hinge reactions have in general kinetic com- 
ponents, i.e., the hinge reactions depend in part at least upon 
the motion. These kinetic components continually change 
direction so that the bearings are subjected to injurious influences 
to prevent which it is sometimes desirable to "balance" the 
rotating system. 

Balancing consists in arranging the rotating bodies on the 
shaft so that the effective forces for the rotating system are in 
equilibrium; then the kinetic components of the hinge reac- 
tions vanish, and the hinge reactions depend only on external 
forces. 



280 ROTATION. [Chap. XII. 

In the following we consider only rotation at constant speed 
(ck = o), and rotating systems consisting of one or more bodies 
which have planes of symmetry perpendicular to the axis. Then 
the resultant effective force for each rotating body is a force 
whose action line is in that plane and passes through the mass- 
centre and the axis, and whose magnitude is mrco (m denoting 
mass of the body, r the distance of its mass-centre from the axis, 
and co the angular velocity of the system). 

Since the centrifugal force which each rotating body exerts 
upon the shaft is equal and opposite to the resultant effective 
force for the body, balance is effected if the centrifugal forces 
(or the reversed effective forces) for all the rotating bodies are 
in equilibrium. This is the usual view of balancing and we shall 
follow it. 

For convenience we will write r instead of r and will often 
say " body" instead of " mass-centre of a body." We will also 
denote both a body and its mass by m. 

276. Balancing of Bodies whose Mass-centres are in the Same 

Plane. — Let m lt m 2 , and m 3 (fig. 
213) represent three bodies whose 
common centre of rotation is O. 
>B Their centrifugal forces equal 
respectively 

m^w 2 , m 2 r 2 co 2 , m 3 r 3 co 2 , 

' 2I3 ' their directions being from to 

the mass-centre of the rotating body in each case. 

Let AB, BC, and CD represent the magnitudes and directions 
of the three centrifugal forces. Then a fourth force acting 
through 0, which would close the force polygon ABCD, would 
balance the three centrifugal forces. This fourth or balancing 
force can be supplied by adding to the system of rotating bodies 
a fourth one whose centre of rotation is 0, whose mass-centre is 
in the direction DA from 0, and whose distance from the axis 
(r) and mass (m) are such that 

mrco 2 = DA (by scale), or mr = DA/co 2 . 
It will be noticed that co is a common term in the forces rep- 





§ III.] APPLICATIONS. 281 

resented by the lines of the force polygon. Hence, if the poly- 
gon for the centrifugal forces closes, it will also close for forces 
parallel to them of magnitudes m^, m 2 r 2 , etc.; so that if AB, 
BC, and CD are drawn parallel to the first three centrifugal 
forces respectively but equal (by scale) to m x r x , m 2 r 2 , m s r 3 , then 
DA represents mr to that same scale. 

277. Balancing Rotating Bodies whose Mass-centres are in 
Different Planes. — The general method consists in finding the re- 
sultant of the centrifugal forces of the rotating bodies and then 
arranging one or more additional bodies so that the centrifugal 
forces due to them will balance that resultant. 

In general, the resultant is not a single force but in all cases 
the forces can be compounded into a force and a couple (see art. 
58 for the method of performing this composition), and as we 
shall see, balance may be effected by means of the addition of 
two bodies the positions of which along the axis may be taken 
arbitrarily. 

Let m x and m 2 (fig. 214) be two bodies rotating about the 
common axis 1 2 , r x and r 2 denoting their distances from the 
axis. Their centrifugal forces are m x r x cu 2 and m 2 r 2 cu 2 respect- 
ively. Let and 0' be taken as centres of rotation of the 
balancing bodies.* 

We replace each one of the given centrifugal forces hy a 
force at and a couple. Thus m x r x co 2 is replaced by F x { = m x r x a) 2 ) 
and a couple whose moment is m x r x oj 2 -a x whose plane is that of 
the centrifugal force and the axis, and whose sense is counter- 
clockwise when viewed from above. This couple is represented 
by the vector C x . The replacing force and couple for the second 
centrifugal force may be similarly described; they are repre- 
sented by F 2 ( = m 2 r 2 aj 2 ) and C 2 (=m 2 r 2 co 2 -a 2 ) respectively. 

Fig. 214(a), being a perspective, does not represent F x , F 2 , 
C x , and C 2 in their true relation; fig. 214(6) does. If ABC is a 
force polygon for F x and F 2 , CA represents the magnitude and 

* It will perhaps occur to the student that the simplest way to balance 
m x and m 2 is to balance each directly by means of a single body rotating 
about #i and 2 respectively. While this may be true for two bodies, 
it is not generally true for more than two. The method here given holds 
for anv number of bodies. 



282 



ROTATION. 



[Chap. XII. 



direction of a force which will balance F t and F 2 , and if abc is 
the "couple polygon" for the couples C t and C 2 , ca is the vector 
of a couple which will balance C x and C 2 . The balancing force 
is represented by F (fig. 214a), and the balancing couple by the 
vector C\ the shaded plane (perpendicular to C) is that of the 
balancing couple. 

The balancing couple can be supplied by the addition of two 
bodies in the shaded plane. Let m! and m f (in the planes through 




Fig. 214. (After Dalby.) 

O and O' and in the shaded plane) be two bodies of equal mass 
and equally distant (r r ) from the axis. The centrifugal force 
due to each is m'r'w 2 , and if 



or 



w'r' = C/co 2 a, 



these two bodies furnish the balancing "centrifugal couple." 

The balancing force can be supplied by the addition of a 
single body. Let m (in the direction F from 0) be a body dis- 
tant r from the axis so that 

mro?—F\ 

then the centrifugal force due to that body balances F x and F 2 . 



§111.] APPLICATIONS. 283 

We have now balanced the system by the addition of three 
bodies, but the two having the same centre of rotation can 
oe replaced by a single one {m"). The centrifugal force due to 
m" must be equivalent to the resultant of the centrifugal forces 
due to the bodies to be replaced (m and m'). Thus, draw AD to 
represent the centrifugal force of m'\ then CD represents the 
centrifugal force of m" , and this replacing body should be placed 
in the direction CD from at such a distance r" that 

m"r"u? = CD, or rn"r f ' = CD/w 2 . 

Thus, balance has been effected by two bodies whose centres of 
rotation are arbitrarily situated along the axis and whose masses 
(m! and m") or distances from the axis (r' and r") may also be 
arbitrarily selected. 

EXAMPLES. 

1. Let m t and m 2 equal 20 and 12 lbs., r x and r 2 equal 6 and 9 
in., and 1 2 equal 15 in., and suppose that w x andw 2 are in the 
same plane with the axis and on the same side of it. Show how 
the two bodies might be balanced. 

2. If the two bodies of ex. 1 are on opposite sides of the axis, 
show how they might be balanced. 

3. Let m lf m 2 , and m 3 equal 20, 12, and 18 lbs., r lt r 2 , and r 3 
equal 6,9, and 12 in., X 2 and 2 3 equal 16 and 18 in. respect- 
ively ,_and suppose that the successive angles between x m x , 2 m 2 , 
and O3W3 equal 90 . Show how to balance the three bodies with 
additional ones rotating about X and 3 . 

278. Pivot and Journal Friction. — Flat Pivot. — Let fig. 215(a) 
represent the flat end of a shaft which is pressed or "thrust" 
against a flat bearing and rotates. Let r denote the radius 
of the end, N the thrust or normal pressure, and / the coeffi- 
cient of friction for the rubbing surfaces. 

Regarding the normal pressure as uniform, its value per unit 
area is N/nr 2 , and the normal pressure on any elementary 
area as that indicated in fig. 215(a) is (N/nr 2 )pdddp while the 
friction on that area is / times that elementary pressure. The 
moment of the elementarv frictional force about the axis of 



284 ROTATION. [Chap. XII 

the shaft is p times the force, and the sum of all such moments 



is 



f(N/nr 2 )£ f 2 *p 2 dpdd = fN%r. 



The resultant friction is not a force but a couple, and hence we 
may regard the actual frictional resistance as a couple whose 
forces equal fN and whose arm is f r. 




Fig. 215. 

Conical Pivot. — Let fig. 215(6) represent a rotating conical 
pivot which is pressed into its step by a thrust P directed along 
the axis. Let r denote the radius of the step, cj> the angle shown, 
and p the normal pressure per unit area regarded as constant. 

The normal pressure on an elementary area dA of the bear- 
ing is pdA and its vertical component is (pdA) sin cj>. Since 
the friction has no vertical component the sum of all the ver- 
tical components of the normal pressure must equal P, hence 

P = p sin <f> fdA = p sin cf> • A = pnr 2 , or p = P/nr 2 . 

Note that the normal pressure per unit area is independent of <j>. 
The frictional force on each element of area dA is j(P/nr 2 )dA , 
and its moment with respect to the axis is p times the force (see 
the figure) ; hence the entire frictional moment equals 

KP/Kf*) fdA.p. 

For simplicity take dA of such form that its horizontal projec- 



§ III.] APPLICATIONS. 285 

tion equals pdpdd (see the figure), i.e., that (dA) sin <j>=*pdpdd. 
Then the above expression can be written 

Journal Friction. — In general it is not known how the nor- 
mal pressure (and hence the friction) varies over the surface of 
a journal. It is customary to compute the "axle" or "journal 
friction ' ' from 

F = f'R, 

in which F denotes the value of a single resistance applied to 
the surface of the journal whose moment about the axis is the 
same as that of the actual frictional resistance, R the resultant 
pressure between journal and bearing, and f a coefficient of 
journal friction. 

The coefficient f is determined from f = F/R, R and F hav- 
ing been experimentally measured. The values of /' and of / 
(for pivots) depend on circumstances, as described below. They 
range from about 0.004 * n the most favorable cases to about 0.0& 
in ordinary lubrication. 

Friction of Lubricated Surfaces * — "The laws which appear to 
express the behavior of well-lubricated surfaces are almost the 
reverse of those of dry surfaces." Thus the frictional resistance 

(1) is almost independent of the pressure with bath lubrica- 

tion . . . ; 

(2) varies directly as the speed for low pressures . . . ; 

(3) depends more upon the temperature than on any other 

condition . . . ; 

(4) with flooded bearings , depends but slightly upon the nature 

of the material of which the surfaces are composed...; 

(5) of rest is enormously greater than the friction of motion... ; 

(6) is least at first, and rapidly increases with the time after 

the two surfaces are brought together. . . . 

* Abridged and quoted from Goodman's "Mechanics Applied to En- 
gineering." Chap. VII of that work is an extensive discussion of the 
subject of Friction. 



286 ROTATION. [Chap. XII. 



EXAMPLES. 

i. Show that the value of the frictional moment on a hollow 
flat pivot is %fN{r 2 z — r^)/{r 2 2 —r x 2 ) % r t and r 2 denoting the inner 
and outer radii respectively of the pivot. 

2. Deduce an expression for the frictional moment on a 
pivot formed of a frustrum of a cone, there being no pressure 
on the lower base of the frustrum. Use notation of fig. 215(6) 
and call the radius of the step r 2 and that of the end of the 
pivot r v 




CHAPTER XIII. 
ANY PLANE MOTION OF A RIGID BODY (RESUMED). 

§ I. General Principles. 

279. The Effective Forces. — Let fig. 216 be a section of the 
moving body parallel to the plane of the motion, and P and P' 
any two points of the body in that section. 

Let a! denote the acceleration of P' ; 
a the angular acceleration of the 

body; 
a) its angular velocity ; 
r the distance P'P ; 
dm " mass of the particle at P. 
According to art. 220, the acceleration of P ^p , 

can be resolved into three components as 

shown in the figure, one being the same as the acceleration of 
P' and the other two being components of the acceleration of 
P relative to P' . Therefore the effective force for the particle 
at P can be resolved into three components whose directions are 
the same as those of the three component accelerations, their 
values being . 

dm -a', dm- remand dm-rw 2 . 

In accordance with art. 221, all the components like dm, -a' may 
be called the translation effective forces and all those like dm • ret 
and dm-reo 2 may be called the rotational effective forces. 

280. Moment of the Effective System. — The moment axis is 
taken perpendicular to the plane of the motion and through any 
point of the body. Let P' (fig. 217) be the point and P'X and 
P'Y be two axes parallel to the plane of the motion and fixed in 
direction. Now the moment of the effective force for the par- 
ticle at P equals the sum of the moments of its components ; the 

287 



288 



ANY PLANE MOTION OF A RIGID BODY. [Chap. XIIL 



moment of dm-roj 2 is zero, that of dm-ra is (dm-ra)r, and that 
of dm • a' equals the sum of the moments of its x and y compo- 
nents (dm-aj and dm-a y ) or — (dm-a x )y and (dm-a y )x re- 
spectively. Hence the moment of the effective force equals 

(dm -ra)r-\- (dm • a y ')x — (dm • a x ')y, 

and the sum of all such moments (for all the particles of the 
body) is 

a I dm • r 2 + a y / dm -x — a x l dm • y, 

or la + mayX — ma x 'y, 

I being the moment of inertia of the body with respect to the 
moment axis, x and y the coordinates of the mass-centre at the 
instant for which the moment is computed. 




Fig. 217. 

This expression for moment simplifies considerably for two 
special moment axes, as follows: 

(a) If the moment axis contains the mass-centre, x and y are 
zero, and the expression for moment reduces to la, I denoting 
the moment of inertia with respect to that axis. 

(b) If the moment axis coincides with the instantaneous axis 
of no acceleration (the line all points of which have at the instant 
no acceleration), a x and a y ' are zero, and the expression for 
moment reduces to la , I denoting the moment of inertia of the 
body with respect to that axis. 

281. Equations of Motion. — Let the motion of the mass-cen- 
tre be referred to a set of fixed axes, x, y, and z, the last being 
perpendicular to the plane of the motion ; also let 



§ I.] GENERAL PRINCIPLES. 289 

a x denote the x acceleration of the mass-centre ; 
d y the y acceleration of the mass-centre ; 
IF X ' ' sum of the x components of the external forces ; 
IFy " " " " y " " " 

IM " " " " moments " " about 

an axis through the mass-centre and perpendicular to 
the plane of the motion. 
As shown in art. 240 

iT^ma*, l¥ y = m* y ; also IM = Ia, . . (1) 
since, according to D'Alembert's principle, the sums of the mo- 
ments of the external and effective forces about any line are 
equal. 

282. Resultant of the Effective System in Important Special 
Cases. — It is assumed in this article that the body is homogene- 
ous and has a plane of symmetry which is parallel to the plane 
of the motion. Imagine the body divided into elementary rods 
perpendicular to the plane of the motion. As explained in art. 
264, the effective force for each of these is one whose action line 
is in the plane of symmetry; hence the effective forces for all 
the rods constitute a coplaner system, its plane being the plane 
of symmetry. 

We proceed to determine the resultants of the effective forces 
corresponding to translational and rotational components of 
the motion. In general, the resultant of each set is a single 
force. Let fig. 218(a) represent the section of symmetry of the 
body, C the mass-centre and P' any other point of the section. 
In addition to the notation of the foregoing articles, let 

r denote the distance from C to P' ; 

a' the acceleration of C relative to P', i.e., its acceleration in the 

rotational component; 
k the radius of gyration of the body with respect to the 

axis through P' perpendicular to the plane of the motion. 

(a) Regarding the motion as resolved into a rotation about 
the axis through P' and the corresponding translation. — Accord- 
ing to art. 244, the resultant of the translational effective forces 
equals ma' and acts in the direction of a' through the mass- 



290 



ANY PLANE MOTION OF A RIGID BODY. [Chap. XII. 



centre as shown. According to art. 262, the resultant of the 
rotational effective forces equals ma' and acts in the direction 
of a' through a point Q in the line P'C such that P'Q equals 
k 2 /r. This last force may be resolved into two components mm 
and tnrco 2 , as represented, fig. 218(a). 

(b) Regarding the motion as resolved into a rotation about 
the axis through the mass-centre and the corresponding trans- 
lation (P' coincides with C). — The resultant of the translational 
effective forces is a force equal to ma and acts in the direction of 
a through the mass-centre as shown in fig. 218(6). Accord- 
ing to art. 262, the_resultant of the rotational forces is a couple 
whose moment is la. 




Fig. 218. 

If the acceleration of the mass-centre and the angular accel- 
eration of the body are both zero, the resultant of the effective 
forces vanishes. 



§ II. Applications. 

283. Determination of the Motion. — The general method 
consists in writing the equations of motion for the case in 
hand and deducing from them the value of the angular accelera- 
tion of the body and that of the acceleration of a point of it. 
The method is further explained in the solution of some of the 
following 

EXAMPLES. 

i. A homogeneous cylinder rolls without slipping down an 
inclined plane. Determine the motion. 




§ II. ] /IP PLICA TIONS. 2 9 1 

Solution: Let W denote the weight of the cylinder, and m 
its mass ; also let F and TV denote the com- 
ponents of the reaction of the plane along 
and perpendicular to its surface (fig. 219).* 

The external forces acting on the cyl- 
inder are F, N, and W; hence with coor- 
dinate axes as shown, eqs. (1), art. 281 be- 

come Fig. 219. 

IF X = — W sin <f>—F = ma Xi IF y = N — W cos <j> = ma yi 

IM = Fr = Ta. 

Evidently a y = o, hence a = a x . Now v = rcu (see ex. 2, art. 220); 
hence a = ra. This equation and the first and third above deter- 
mine a and a. We find that (since I = %mr 2 ) 

a=f(gsin <f>)/r, 
a = fgsin </>. 

We may also determine the reaction of the plane ; from the 
first equation 

F = %W sin cf> and from the second, N = W cos <j>. 

2. Show that there will be no slipping between the cylinder 
and plane if | tan <j> is greater than the coefficient of static 
friction. 

3. If in ex. 1 <f> = $o° and the coefficients of static and kinetic 
friction are respectively 0.25 and 0.2, determine the angular 
acceleration of the cylinder and the linear acceleration of its 
mass-centre. 

4. A homogeneous sphere rests on a smooth horizontal sur- 
face and a horizontal force (P) is applied by means of a string 
wrapped about it along the horizontal great circle. Determine 
the angular acceleration of the sphere and the acceleration of 
its mass-centre. 

* We assume in this and the following examples on rolling that the 
rolling body and the surface on which it rolls do not distort each other, 
thus leaving a point or line of contact; then there is no "rolling resist- 
ance" (art. 285). 



292 ANY PLANE MOTION OF A RIGID BODY. [Chap. XIII. 

Solution : The external forces acting on the sphere are P, its 
weight (W), and the resistance (N) of the surface. Equations 
(i) of art. 281 become (see fig. 220), 

2F x =P=ma x , IF y = o = ma y , IM = Pr=Ia. 

From the last equation, a=Pr/I, and from the first d x = P/m. 
Since a y = o, a = a x = P/m. 

5. Solve ex. 4, supposing that the plane is rough, its coeffi- 
cient of kinetic friction being /. 

6. Fig. 221 represents a wheel rolling on a horizontal sur- 
face. If the distance of its centre of gravity 
from the centre is c, and its radius of gyration 
with respect to an axis through its centre and 
perpendicular to the plane of the motion is k, 
show that 




— gc sin 6 = rc sin 6- w 2 + (r 2 — 2rc + k 2 )a. 

284. Kinetic Reactions. — In the following 
examples the motion of the body under consid- 
eration is given and the unknown quantities are 
forces . These are determined by use of the equa- 
tions of motion, but sometimes other forms of the equations are 
more convenient. These other forms are obtained, like those 
of art. 281, by D'Alembert's principle, i.e., by /^T~~^\ 
expressing algebraically the fact that the ex- / :r \ 
ternal forces acting on a body are equivalent / i \ 

to the resultant effective force for it. This I f^\ c / 

equivalence can be expressed in many ways; \. jw| -S 
thus, to deduce the equations of art. 281, we 
wrote two resolution equations and one mo- -^ IG - 221, 

ment equation with moment-axis through the mass-centre 
and perpendicular to the plane of the motion. But we may 
write one resolution and two moment equations or three moment 
equations. The first example below is solved from two sets of 
equations, thus illustrating different forms of the equations of 
motion. 



§H.] 



APPLICATIONS. 



293 



EXAMPLES. 

i. A wheel whose mass-centre is not in its geometrical axis 
is rolled along a horizontal roadway at a constant speed by means 
of a horizontal force (P) applied at its centre. Determine the 
value of P and the reaction of the roadway for any position of 
the wheel. 

Solutions: (1) By use of the equations of art. 281. Let fig. 
222 represent the wheel, P' being its centre and C its mass-cen- 
tre, and let F and N denote the hori- 
zontal and vertical components of the 
reaction of the roadway. The exter- 
nal forces acting on the wheel are W, 
P, F, and N; hence eqs. (1), art. 281, 
become 



2F X =P-F = 

IF y = N-W 



■ma. 



ma, 



IM=F(r+c sin 0) 

— Pc sin d = o. 



Nc cos 6 



(1) 
(2) 

(3) 




Fig. 222. 



Now the acceleration of C (a) equals the vector sum of the accel- 
eration of P' and that of C relative to P' (art. 196). Since the 
wheel rolls uniformly, the acceleration of P' is zero and that 
of C relative to P is ceo 2 in the direction CP' ; hence 

a = cco 2 directed along CP' . 

Also a x = ceo 2 cos 6 and a y = — coo 2 sin d, 

and eqs. (1), (2), and (3) may be written 

F = P — mcco 2 cos 6, N = W— mcco 2 sin 0, and 

(P — mcco 2 cos 6) (r + c sin 6) — ( W — mcco 2 sin d) c cos O—Pc sin 6 — o. 

From these we find that 

P = (Wc/r + mcco 2 ) cos d, 
F = W(c/r) cos 0, 
and N = W — mcco 2 sin 0. 

(2) The three components of the resultant effective force as 
described in art. 282, case (a), are ma' , mra, and mrco 2 . P' 



294 ANY PLANE MOTION OF A RIGID BODY. [Chap. XIII. 

(fig. 2 1 8a) being chosen at the centre of the wheel, a' is zero; 
and since a is zero, the resultant effective force is mra> 2 = mcoj 2 
directed from C to P'. Now by D'Alembert's principle F, N, 
P, and W are equivalent to mew 2 , or they are in equilibrium with 
mcoj 2 reversed. Hence 

IM A =Pr — Wc cos 6 — mcu?r sin = o, 
IMp, = Fr - Wc cos = o, 
IF y = N-W-mcoj 2 sin0 = o. 

From these (without elimination) we get the same values for P, 
F, and N as were found in (i). 

The kinetic reaction (on the wheel) is vertical and equals 
mcco 2 sin ; it acts downwards for values of 6 between o° and 
i8o° (C is above P') and upwards for values of 6 between i8o° 
and 360 (C is below P'). Otherwise stated, when C is above 
P', N is less than W, and when C is below P', N is greater than 
W. The kinetic component of the reaction N is called " ham- 
mer blow" in locomotive parlance. 

2. Show that the wheel of ex. 1 lifts from the roadway in a 
certain position if the speed of the centre of the wheel is greater 
than r\/g/c. 

3. For which portion of C is N a maximum, and what is that 
value if the velocity is slightly less than r\/g/c. 

4. A steam-engine connecting-rod with no piston-rod attached 
is drawn by the crank. It is required to determine the kinetic 
reactions at the ends of the rod in any position. 

Solution: Let AB and P'B (fig. 223) be the crank and rod 
respectively, C being the mass-centre of the latter. Let r de- 
note the length CP', a the acceleration of P' r m the mass of the 
rod, a and (o its angular acceleration and velocity respectively, 
and k the radius of gyration of the rod with respect to a line per- 
pendicular to the plane of the motion at P'. 

We first determine the resultant effective force (R) for the 
rod. Its three components as given by art. 282 are 

ma, mra, and mrco 2 , 
acting as shown in the figure, the distance of Q from P' being 



§11.] 



APPLICATIONS. 



95 



k 2 /r.. Supposing these to have been computed, R can be readily 
found graphically (the construction is not shown). 

Let V and P denote the reactions on the rod at P f and B 
respectively; the "guide-rods" being supposed frictionless, V 




Fig. 223. 

is vertical. Since V, P, and R reversed are in equilibrium, their 
action lines intersect in a point. So produce R to intersect V 
and join that intersection with B ; this line is the action line 
of P. To determine the values of V and P, we draw a force 
triangle for V, P, and R reversed. Thus, lay off MN to repre- 
sent the magnitude of R, and from TV draw a vertical line to inter- 
sect the action line of P, marking that point 0; NO and OM 
respectively represent the magnitudes and directions of V and P. 

5 . Let r and c denote the lengths of the connecting-rod and 
crank respectively, and a> the angular velocity of the crank 
assumed constant. Take r/c equal' to four and compute as in 
ex. 4 the values of V and P when (see fig. 223) equals o°, 30 , 
6o°, oo°, 120 , 150 , and 180 . Also plot, the values of V at the 
corresponding positions of P' and represent each P, scaling it 
from the corresponding position of B\ then draw smooth 
curves, joining the ends of the vectors representing V and P 
thus drawn. 

(For a method of computing a, a, and <o, see art. 181 and 
ex. 1, art. 219. Notice that of fig. 154 equals 6 — go° of figs. 
175 and 223.) 

285. Rolling Resistance. — Let fig. 224 represent a wheel or 
cylinder rolling upon a horizontal surface or roadway at con- 
stant speed. Let W denote its weight, P, applied as shown 



296 ANY PLANE MOTION OF A RIGID BODY. [Chap. XIIL 

the force required to maintain the constant speed, and R 

the resultant reaction of the roadway. Since the accelera- 

^ — .J. tion of the mass-centre and the angular accel- 

f ^\ eration of the wheel are zero, the resultant 

/ \ effective force for the wheel is zero (art. 282); 

L— r ^ < J hence the external forces W, P, and R are in 

1 1 p I . 

\ I equilibrium, and R must act through the cen- 

\ / tre and be inclined as shown. 

^^ I ^^ The horizontal component of the reaction 

/j - ! of the roadway is called " rolling resistance," 

' ' also " rolling friction." The three forces being 

in equilibrium, the horizontal component of R 

equals P, and hence an expression for P is also a value of the 

rolling resistance. If a denotes the distance from the vertical 

through the centre to the intersection of the action line of R and 

the circumference of the wheel, and if a is small (as it is except 

on soft roadways), then approximately 

Pr = Wa, or P = (a/r)W. 

From this equation the rolling resistance in a given case can be 
computed if a is known. 

Like the coefficients of friction, a is determined from exper- 
iment. Thus, suppose that the force P for a given wheel and 
roadway has been measured; that value and the weight and 
radius of the wheel substituted in either preceding equation 
determine a for the case in hand. The distance a is sometimes 
called the "coefficient of rolling resistance." Observe that it is 
not an abstract number like the coefficients of friction, but a 
length. 

Practically no general facts or laws are known concerning 
the coefficient of rolling resistance. The coefficient is usually 
regarded as independent of the weight (if moderate so that no 
permanent deformation of roller or roadway occurs), but there 
is disagreement as to the relation between it and the radius — 
it being held, for example, that a is independent of r, also that 
a varies as Vr. 

The following values are given to afford a notion of the values 
of a in a few cases. 



§ II J APPLICATIONS. 297 

Rollers of elm on an oak track (Coulomb) 0.032 in. 

Iron or steel wheels on iron or steel rails 0.007-0.020 " 

" " " " "wood 0.06-0.10 " 

EXAMPLE. 

If several rollers are used for rolling a heavy body along a 
horizontal roadway, show that their rolling resistance is given 
by W(a' + a")/2r, W denoting the weight of the body, r the 
radii of the rollers, and a' and a" their "coefficients of rolling 
resistance" for the surfaces at which the rolling occurs. 



CHAPTER XIV. 



WORK AND ENERGY. 



§ I. Work. 

286. Work Defined. — Work is said to be done upon a body 
by a force when the application point is displaced so that it has 
a component along the action line of the force. 

If the force is constant in direction, the projection of the 
displacement on the action line of the force is called effective 
displacement. If the force changes its direction during a displace- 
ment of its application point, then the projection of the dis- 
placement which occurs during an element of time on the 
corresponding action line of the force is called the effective 
displacement for that elementary interval. 

287. Expressions for Work Done by a Force. — I. The force is 
constant in magnitude and direction. The 
amount of work done is measured by the 
product of the force and the effective dis- 
placement. Thus, if AB (fig. 225) is a dis- 
placement of the application point of a force 
F, and if <f> denotes the angle between AB 
and F, and w the work done by F, then 




Fig. 225. 
If cj) = o, w = FAB 



w = F(AB cos <j>) = (F cos </>)AB. . (1) 

if <j) = go°, w = o. 

Observe that in the last form of (1), the expression for work 
is the product of the component of the force along the displace- 
ment and the displacement. 

II. The force varies in Magnitude or Direction or both. Let 
AB (fig. 226) be a portion of the path of the application point of 
the force F, P being any intermediate position, <j> the angle be- 
tween F and the tangent to the path at P, and s the distance 
of P from some fixed origin in the path, it being measured posi- 

298 




§ I.] WORK. 2 99 

tively in the direction of the motion. Then if dw denotes the 
work done by the force while its applica- 
tion point describes an elementary portion 
of the path ds including P, 

dw = F ds cos (j) ; 

and if w denotes the work done by F dur- 
ing the displacement AB in which s 
changes from s' to s", 

w=f s FdscoS(j)=rF t ds (2) 

If <f> = o, Ft = F; if <£ = 9o°, F t = o and w = o. 

288. Sign of a Work. — It is convenient to give sign to the 
work done by a force. The rule is as follows: A work is re- 
garded as positive or negative according as the effective dis- 
placement agrees with or is opposed to the force in sense. It 
must be remembered that a displacement, and hence its pro- 
jection also, is a vector quantity. 

If the angle <f> is always taken as in figs. 225 and 226, i.e., be- 
tween the portions of the lines representing the force and the 
displacements toward which the arrows point, then the expres- 
sions for w in the preceding article give the correct sign of the 
work. 

289. Unit of Work. — Equations (1) and (2), art. 287, imply 
as unit the work done by a unit force ' ' acting through ' ' unit dis- 
tance. The value of the unit hence depends on the units used 
for force and distance. Thus we have, corresponding to the 
pound and foot, the foot-pound unit of work, to the kilogram 
and meter, the meter-kilogram, to the dyne and centimeter, the 
dyne-centimeter, etc. The last-named unit has also a special 
name, erg* 

290. Work Diagram. — If values of F and 5 be plotted on two 
rectangular axes (see fig. 227) for all positions of the application 
point of the force F, the curve joining the plotted points might 
be called a "tangential force-space (or F r s) curve." The por- 

* For dimensions of a unit work, see Appendix C 



3°° 



WORK AND ENERGY. 



[Chap. XIV. 




tion of the figure between the curve, the s axis, and any two 
ordinates is called a work diagram. 

Proposition. — The area of a work diagram represents the work 
done by the force during the displace- 
ment corresponding to the bounding 
ordinates. 

Proof: According to eq. (2), art. 287, 

w = I f Ftds. Since F t and 5 also denote 

— coordinates of points on the F r s curve 

Fig. 227. (fig. 227), the area of the work diagram 

is / F t ds } as is shown in works on calculus. Hence the area 

(according to some scale) equals w. Obviously the scale accord- 
ing to which the area of a work diagram is to be interpreted 
depends on the scales used in representing F t and s. Thus if 
one-inch ordinates and abscissas represent 100 lbs. and 10 ft., 
respectively, one square inch of area represents 1000 ft. -lbs. of 
work. 

Since the area of a work diagram equals the product of the 
average ordinate and the base, the work done by a force equals 
the average value of the tangential component and the length 
of the path described by the application point. If the F r s curve 
is straight, the average value of F t is the mean of the initial and 
final values and the computation of the work is simple. 

EXAMPLES. 

1. Fig. 228 represents a body on a horizontal surface to which 
two forces (P and Q) are applied as shown. Compute the work 



— *i 



r4 



w 



P=1001bs 1 . 



s =o F r s =st% 



In 
Fig. 228. 



done by these forces, the weight, and the resistance of the 
surface while the body moves from A to B (5 ft.), Q being a vari- 
able force so that Q (in lbs.) = 25 (in ft.) and the friction 20 lbs. 
Solution: Since the displacements of the application points 



§I.J 



IVORK. 



301 



of W and N are normal to the action lines of the forces, W and 
N do no work. Since F and P are constant in magnitude and 
direction, we use eq. (1) art. 287 for computing the work done by 
them; for P, <j> is zero and for F, <j> is 180 , hence 

the work done by P= 100 • 5 cos = 500 ft. -lbs., and 
' ^=130-5 cos i8o°= — 150 ft. -lbs. 
Since Q varies in magnitude we use eq. (2) art. 287, and cj> 
being 180 , the expression for work done by Q is 



i:'q 



cos 180 ds 



=-/: 



2s ds= —25 ft. -lbs. 



This value can be readily computed from the average vauie of 
Q which is, since Q varies uniformly with s, the mean of its 
initial and final values; these are o and 10 lbs. respectively. 
Hence the average value of Q is 5 lbs., and as it acts through 
5 ft., the amount of work done by Q equals 5X5 = 25 ft. -lbs. 
The sign of the work is negative because the senses of the force 
and displacement are opposite. 

2. Fig. 229 represents a body upon an inclined plane to 
which two forces (P and Q) are applied as shown. Compute 
the works done by them, the weight, and the reaction of the 
plane while the body moves from A to B (10 ft.), the frictional 
resistance of the plane and the weight being 10 and 100 lbs. 
respectively. Arts. Total work done= —26.8 ft. -lbs. 




Fig. 229. 




Scales: Horizontal, 1'—%'; Vertical, 1"= 500,000 lbs. 

Fig. 230. 



3. In punching a hole (2 in. diameter) in a certain iron plate 
(ij in. thick), the pressure (F) between punch and plate varied 
as the ordinates to the curve of fig. 230 (the initial values of F 
are at the left). Estimate the area of the work diagram and 



3° 2 WORK AND ENERGY. [Chap. XIV. 

the amount of work done by the punch on the plate in the oper- 
ation. Also compute the work from the average value of the 
force. 

4. One end of an elastic cord whose natural length is 10 ft. 
is fastened to a body on a horizontal surface and the other end 
to a fixed point in the surface 20 ft. from the body. The ten- 
sion in the cord is observed to be 30 lbs. When the body is 
released it moves toward the fixed point. Draw the work dia- 
gram for the tension in the cord and determine how much work 
is done by the tension in the first and second 5 feet of the dis- 
placement. 

5. Suppose that a gas expands behind a piston in a cylinder 
according to the law pv = C, C being a constant, p the gas pres- 
sure per unit area, and v the volume of the gas. Show that the 
work done by the gas on the piston in an expansion from a 
volume v t to a volume v 2 equals C log £ (^ 2 / , y 1 ). 

6. Show that the work of a central force (one always directed 
toward a fixed point) in any displacement of its application 

point equals — / 2 P dr, in which P denotes the general value of 

Jr\ 

the force, i.e., its value when its application point is any dis- 
tance r from the fixed point and r 1 and r 2 denote the values 
of r at the beginning and end of the displacement respect- 
ively. 

Solution: Let C, fig. 231, be the fixed point toward which P 

acts and OAB the path of the 

j\ _ t application point of P. The 

^-'" 1 ~ value of the work done by P is 

^^^ \l given by eq. (2) art. 287. Since 

G^^~*~~~ r p f-f -^ (see the figure) dr = —ds cos <f>, 

^^Z~ " ^Jf ^ e vauie °^ tn e work as given 

*^*r^ /I by eq. (2) reduces to that 

"b/W- given above - 

7. Suppose that the body 

described in ex. 4 is moved (after 
the cord is attached to it as described) so that the point of ap- 
plication of the cord moves in the circumference of a circle 
whose diameter is the cord in its first position. Compute the 



§ I.] WORK. 3°3 

work done by the tension while the application point describes 
the first quarter circle. 

8. How much work is done by the cord up to the instant 
when it resumes its natural length? 

291. Work Done by Gravity Upon a Body in Any Motion. — 
Proposition. — The work done by gravity upon a body in any 
motion equals the product of its weight and the vertical distance 
described by the centre of gravity, and the work is positive or 
negative according as the centre of gravity has descended or 
ascended. 

Proof: Let w lt w 2 , etc., denote the weights of the particles 



A 



\ / 11 w,> 



y..\ 



v* 



of the body, y/, y 2 , etc., their distances 
above some datum plane (below which 
the body does not descend) at the begin- 
ning of the motion, and y" t y 2 ", etc., 
their distances above that plane at the 
end of the motion (see fig. 232 where 
a' a" is the path of the first particle, FlG ' * 32 - 

b'b" that of the second, etc.). Also let W denote the weight 
of the body and y' and y" the initial and final heights of its 
centre of gravity above the plane. Then the sum of the works 
done by gravity on all the particles is 

*>t(yi"-~yi')+v>Jy%"-y*')+ ■ ■ ■ =(ft>&i"+u>& 2 "+ • • • ) 

-(WJ/+W23//+ . . .) 

According to art. 64, 

w^" +w 2 y 2 " + . . . =\Vy" and {w x yj + w 2 y 2 ' + . . . ) = Wy': 
hence the sum of the works done on all the particles equals 

Wy" - Wy' = W(y" -y'). q.e.d. 

292. Work Done by Concurrent Forces and by Their Result- 
ant. — Proposition. — The work done by any number of concur- 
rent forces in a displacement of their application point equals 
that done by their resultant in that displacement.* 

Proof: Let F t , F 2 , etc., denote the forces, R their resultant, 
and cj)^ (f> 2 , etc., and (f> respectively the angles which the forces 

* It is assumed that the forces and their resultant have a common 
application point. 



304- WORK AND ENERGY. [Chap. XIV. 

and their resultant make with the tangent to the path of the 
application point (taken as explained in art. 288). Then ac- 
cording to art. 36, 

R cos <fi=F 1 cos <j) x + F 2 cos (j> 2 + .... 

Therefore, R cos <f>-ds = F 1 cos <f> 1 -ds+F 2 cos <f> 2 -ds + . . . , 

and / R cos <j> • ds = / F t cos <j> x *ds+ F 2 cos <f> 2 -ds-\- ... ; 

hence, etc. 

293. Work Done by a Pair of Equal, Opposite, and Collinear 
Forces. — Suppose that A and B (fig. 233) are the application 



Fig. 233. 

points of the forces at any instant during the motion and let P 
denote the value of the forces then. Also let %' and y' denote 
the coordinates of A and x" and y those of B, and suppose that 
A moves from A 1 to A 2 and B from B± to B 2 , i.e., assume for sim- 
plicity that the displacements are coplanar. The discussion can 
be easily extended to include non-coplanar displacements. 

The work done by each force equals the sum of the works 
done by its x and y components; for the force P acting on A 
these are 

-fp cos d-dx f and -fPsind-dy', 

and for the force P acting on B they are 

fp cos 6 • dx" and fp sin • dy". 

The work done by both forces equals 

/P[cos ■ {dx" - dx') + sin • {dy" - dy')\ 

It is plain from the figure that 

r 2 = (x"-x') 2 + {y"_-y') 2 , 



§ I.] WORK. 305 

hence r dr = (x" -x')(dx" -dx') + (y" -y'){dy" -dy'), 

or dr = cos • (dx" - dx') + sin • {dy" -dy'). 

Substituting according to this relation in the expression for 
total work we find that the latter becomes 

*r 2 






P dr (when the force P on A acts from B to .4), but 

— I 2 P dr (when the force P on .4 acts from A to B), 

as will be seen by changing the arrows on P in the figure and 
making the necessary changes in the discussion. 

If the distance between the application points of the forces 
does not change during the displacement, dr = o, and the work 
done by the pair of forces equals zero. If P depends only on r, 
then the work done by P depends only on the initial and final 
values of r and not at all on the way in which r changes during 
the displacement. 

EXAMPLE. 

How much work is done by the steam in one cylinder of a 
locomotive during one stroke of the piston ? 

Solution: Consider a forward stroke of the piston and let P 
denote the pressure on the piston when its distance from the 
rear end of the cylinder is r. Then the work done by the pres- 
sures on the piston and rear end of the cylinder in one stroke 
equals 



c 



Pdr = P a (r 2 -r 1 )=P a s, 

r x and r 2 denoting the values of r at the beginning and end of the 
stroke, P a the average value of the steam pressure, and 5 the 
length of stroke. 

This value of the work might also be obtained by computing 
the work done by each pressure separately. Thus let R denote 
the radius of the driving-wheels, then the distance through which 
the locomotive moves in one stroke equals xR, and supposing 
that the locomotive is running forward, the work done by the 
steam on the rear end of the cylinder equals —P a nR, and that 
done on the piston equals P a (?:R+s); hence the work done by 
both pressures equals P a s. 



3°6 WORK AND ENERGY. [Chap. XIV. 

294. Work Done by a Body Against a Force. — It is con- 
venient to use the expression "work done by a body against a 
force" applied to it; we mean by it the negative of the work 
done upon the body by the force. Thus if a body weighing 10 
lbs. is made to rise 5 ft., gravity does —50 ft. -lbs. of work upor 
it and the body does +50 ft. -lbs. of work against gravity. 

In accordance with the above, when the sense of a force and 
that of effective displacement of its application point are oppo- 
site, the work done by the body is positive ; when they are the 
same, the work done by the body is negative. 

§ II. Energy. 

295. Energy Defined. — When the state or condition of a body 
is such that it can do work against forces applied to it, the body 
is said to possess energy. A stretched spring can do work against 
forces applied to it if they are such that it may contract ; a body 
in motion can do work against an applied force which tends to 
stop it. The spring and the body therefore possess energy. 

The amount of energy possessed by a body at any instant is 
the amount of work which it could do against applied forces 
while its state or condition changes from that of the instant to 
an assumed standard state or condition. The meaning of the 
standard condition is explained in subsequent articles. 

The unit of energy must, in accordance with the above, be 
the same as the unit of work. 

296. Kinetic Energy Defined. — Energy is classified into 
kinds depending on the state or condition of the body in virtue 
of which it has energy. Kinetic energy of a body is energy 
which it has by virtue of its velocity. 

297. Kinetic Energy of a Particle. — The amount of kinetic 
energy possessed by a particle at any instant is the work which 
it could do while the velocity changes from its value at that in- 
stant to some other value taken as a standard. It is customary 
to take zero velocity as the standard one; this being understood, 
we may say that the amount of kinetic energy possessed by a 
particle is the work which it can do in " giving up its velocity." 

Proposition. — The kinetic energy of a particle whose mass 
and velocity are m and v respectively equals tynv 2 . 



§ II.] ENERGY. 307 

Pruof : Let R denote the resultant of all the forces acting on 
the particle while it "gives tip its velocity." Let A and B de- 
note the beginning and end of the path, and s r its length. Then, 
according to arts. 292 and 287, the work (w) done by all the 
forces on the particle in the motion from A to B is given by 



w 



= f'R t ds. 



Hence the work done by the particle against the forces, or the 
kinetic energy (E) of the particle, is given by 

E=-f S 'R4s. 
Now Rt = mat — m dv/dt (see art. 236); 

hence E= — I mv dv = %mv 2 . q.e.d. 

298. Kinetic Energy of any System of Particles. — The kinetic 
energy of a system of particles equals the sum of the kinetic 
energies of the separate particles. If m and v denote the mass 
and velocity respectively of any particle of a system, and E the 
kinetic energy of the system, 

E=42mv 2 (1) 

I. Translating Body. — In this case all particles have at each 
instant equal velocities, hence 

or, if M denote the mass of the body, the kinetic energy E is 
given by 

E = JMv a (2) 

II. Rotating Body. — In this case the velocity of any particle 
of the body equals the product of its distance from the axis of 
rotation and the angular velocity of the body (art. 216). Let r 
denote the distance of any particle from the axis and oj the 
angular velocity of the body; then the value of the kinetic 
energy is 

%Jm(ra>) 2 = \i*P2mr 2 . 

Now Imr 2 is the moment of inertia of the rotating body with 



3o8 WORK AND ENERGY. [Chap. XIV. 

respect to the axis of rotation; if / be used to denote this quan- 
tity, the kinetic energy is given by 

E = il« 2 (3) 

III. Body having any Plane Motion. — In this case the state 
of the motion at any instant may be regarded as rotational (see 
arts. 222 and 223). As shown in art. 223, the velocity of any 
particle of the body at any instant equals the product of its dis- 
tance from the line which is the axis of rotation at that instant 
(instantaneous axis) and the angular velocity of the body. The 
reasoning in Case II applies here if the word "instantaneous " is 
inserted before the word " axis " ; then \Ioj 2 is the expression for 
the kinetic energy of a body having any plane motion, / being 
the moment of inertia with respect to the instantaneous axis. 

Since the instantaneous axis in general moves about in the 
moving body, the expression above is not always convenient to 
apply, and another, though not so simple in form, is simpler in 
its application. This expression may be deduced as follows: 
In addition to the notation employed in the preceding, let I 
denote the moment of inertia of the body with respect to a cen- 
tral axis perpendicular to the plane of the motion, v the velocity 
of the mass-centre at the instant considered, r the distance be- 
tween the mass-centre and the instantaneous axis, and M the 
mass of the body. Then 

I = T+Mr 2 (art. 256) and v=ra>; 
hence tfw 2 = \Tu 2 + %M? 2 to 2 , 
or E = ilW£Mv 2 (4) 

Now \I(u 2 is the kinetic energy which the body would have if 
rotating about a fixed axis through its mass-centre with an angu- 
lar velocity co, and \Mv 2 is the kinetic energy which it would 
have if translating with a velocity v. Hence the kinetic energy 
of a body having any plane motion is regarded as consisting 
of two parts, and they are called rotational and translational. 

EXAMPLES. 

1. Express the kinetic energy of a body weighing 1.5 tons 
and moving at a speed of 60 mi.-per-hr. in ft. -lbs. 



§ II.] ENERGY. 3°9 

2. Express the kinetic energy of a cylindrical disk weighing 
3225 lbs. and rotating at an angular velocity of 300 rev.-per-min. 

3. Show that the kinetic energy of a rotating body equals 
\Mv k 2 , M denoting the mass of the body and v& the velocity 
of a point of it whose distance from the axis of rotation equals 
the radius of gyration of the body with respect to that axis. 

4. What is the kinetic energy of a homogeneous right cir- 
cular cylinder which rolls so that the speed of its mass-centre 
is v, its mass being Ml Ans. \Mv 2 . 

299. Potential Energy Defined. — A body may possess energy 
which is not due to velocity. Thus two mutually attracting 
bodies can do work against forces applied to either or both if 
allowed to move so that they approach each other; and as 
mentioned in art. 295, a compressed or stretched spring can do 
work against applied forces if permitted to resume its natural 
length. The "change of condhV'on or state" in the first case is 
a change in configuration, i.e., a change in the positions of the 
bodies relative to each other, and in the second case, if we con- 
ceive of the spring as consisting of discrete particles, the change 
is also one in configuration. 

Energy of a system of particles dependent on configuration 
of the system is called energy of configuration and, more commonly , 
potential energy. 

300. The Amount of Potential Energy possessed by a system 
in any configuration is the work which it can do in passing from 
that configuration to any other taken as a standard, it being 
understood that no other change of condition takes place. The 
standard configuration maybe chosen at pleasure, but it is con- 
venient to so select it that in all other configurations considered 
the potential energy is positive. 

Proposition. — The potential energy of a system in any con- 
figuration equals the amount of work done by the internal forces 
during the change to the standard configuration. 

Proof: To determine the potential energy we are to compute 
the work done against the external forces while the system 
passes to the standard configuration, no other change of con- 
dition (as velocity) of the particles taking place; hence in the 
passage to the standard configuration there is no change in the 



3io 



WORK AND ENERGY. 



[Chap. XIV. 



kinetic energy of the system. It is shown in art. 306 that the 
sum of the works done by the internal and the external forces 
during any change of configuration equals the increment in the 
kinetic energy of the system. Since in the case in hand there 
is no change in kinetic energy, the sum of the works done by 
the internal "and external forces equals zero. Denoting the 
internal and external works w { by and w e respectively, 

Wi+w e = o, or w%— —w e . 

Now — w e is the work done by the system against the external 
forces during the passage to the standard condition, i.e., the 
potential energy of the system in its initial configuration ; hence 
the last equation asserts the truth of the proposition. 

301. Potential Energy of a System Not Always a Definite 
Quantity. — The amount of work done by the internal forces dur- 
ing a change of configuration (and hence the potential energy 




Fig. 234. 

of the system) may or may not depend upon the way in which 
the change of configuration takes place. This is known to be 
true from direct experience, but it can also be proved. Thus, 
let A, B, and C (fig. 234) represent three bodies which may slide 
about on a table ; imagine them connected by elastic cords as 
shown, and consider the three bodies, the table, and the earth 
as a system. The cords are introduced merely as a means of 
applying certain forces to the bodies, but the forces are to be 
thought of as exerted by one body directly upon another. Let 
A , B , and C represent a selected standard configuration of the 



§11.] ENERGY. 3 11 

bodies and A,B, and C some other one ; also , let F lt F 2 , and F 3 be 
such external forces acting upon them that the passage to the 
standard position is without change of velocity. 

Now the only internal forces in the system which do work 
during the motion are the pulls exerted by the bodies upon each 
other and by the frictions (if any) between them and the table. 
As shown in art. 293, the work done by the pulls exerted by any 
two of the bodies on each other does not depend on how the 
final positions are reached. The work done by the frictions, 
however, does depend on the manner of the motion; thus, if we 
suppose the friction on any body to be constant in value, the 
work done by that force equals the negative product of the force 
and the length of the path described by that body. Hence, in 
the system of bodies under consideration, the total work done by 
the internal forces depends on the way in which the change of 
configuration takes place and the potential energy is not a 
definite quantity. If, however, there is no friction, the work 
done by the internal forces is independent of the way in which 
the configuration takes place, and the potential energy is a 
definite quantity. 

302. Conservative Systems. — If the work done by the internal 
forces of a system during any change of configuration is inde- 
pendent of the way in which the change is made, the system is 
called conservative, and those internal force-pairs (action and 
reaction) whose work does not depend upon the way in which 
the change takes place are also called conservative. The poten- 
tial energy of such a system in any definite configuration is a 
definite quantity. 

It is characteristic of conservative forces that they are inde- 
pendent of the velocity of the particles on which they act. We 
consider only conservative forces which act along the line join- 
ing the particles between which they act and whose magnitudes 
depend on the distance between the particles. That such forces 
are conservative forces follows from art. 293. 

303. Non-Conservative Systems. — If the total work done by 
the internal forces depends on the manner of the change of con- 
figuration, the system is called non-conservative and those inter- 
nal force-pairs whose work depends on the manner of the change 



312 WORK AND ENERGY. [Chap. XIV. 

are also called non-conservative. The potential energy of a 
non-conservative system in any definite configuration is not a 
definite quantity. 

It is a characteristic of non-conservative forces that their 
magnitudes or directions (or both) depend upon the velocity of 
the particles to which they are applied. Friction is the only 
one of this class herein considered. 

304. Localization of Potential Energy. — Unlike the kinetic 
energy, the potential energy of a system cannot always be 
localized in detail, i.e., we cannot in all cases assign to parts of 
the system certain definite parts of its potential energy. As an 
example, consider two bodies A and B of the illustration in art. 
301, and, for simplicity, neglect friction. As previously ex- 
plained, the bodies can do a definite amount of work against 
external forces in passing to their standard positions, A B (fig. 
234), but the amount of work which each can do depends upon 
the way in which the passage is made. Thus, suppose that they 
move to their standard positions successively and that in one 
passage A moves first and in the other B moves first. Now the 
work which A can do in each case equals the work which the 
internal force acting on A does, and this work has different 
values in the two passages. For, let P denote the internal force 
acting on A (for simplicity assumed constant) ; then (see ex. 6, art. 
290) the work done by P in the first case equals ±P(A B — AB), 
and in the second case it equals ±P(A B — AB ); these values 
are in general unequal. 

If, however, one of the bodies is always in its standard posi- 
tion, the potential energy of the system is rightly ascribed to the 
other. Thus, a magnet and a piece of soft iron attracting each 
other possess potential energy if separated, and if the magnet is 
regarded as fixed the energy is possessed by the iron. Simi- 
larly, the earth and an elevated body considered as a system 
possess potential energy, but it is practically necessary to regard 
the earth as fixed and hence to ascribe the energy to the elevated 
body. 

305. Other Forms of Energy. — Kinetic and potential energies 
are often called mechanical energy. It is the opinion of some 
that all energy is mechanical, and some think that it is all 



§11.] ENERGY. 313 

kinetic. Whether either of these views be correct, it is practi- 
cally necessary to recognize other forms. A mere enumeration 
of these with brief remarks is sufficient for the present purpose, 
since we shall deal mostly with energy known to be mechanical. 
Thermal Energy. — A hot body may do work under favorable 
conditions; thus, if such a one is placed in a boiler containing 
water, the water will be heated and a part may be converted 
into steam which may drive a steam-engine, i.e., do work. By 
giving up its heat the hot body has done work, and hence by 
definition (art. 295) it possessed energy in its heated state. Not 
only is this fact well known, but also the fact that a given quan- 
tity of heat represents a definite amount of energy ; the relation 
may be expressed thus : 

one British unit of heat * = 778 + ft. -lbs. 

Based on the molecular hypothesis, the common theory is 
that heat is due to the vibratory motion of molecules, i.e., that 
thermal energy is kinetic. 

Chemical Energy. — Many substances combine chemically and 
their combination gives evidence that they possessed energy. 
Thus, coal and oxygen combine and produce heat which, as we 
have seen, is a form of energy. We rightly say, therefore, the 
coal and oxygen before combination possessed energy. 

Based on the molecular hypothesis, the theory of chemical 
energy in cases where heat is generated in the chemical com- 
bination is that internal (molecular) forces of the substances 
do work during the combination, and hence (see art. 306) in- 
crease the kinetic energy of the molecules. According to this 
explanation the energy before combination is potential and after 
kinetic. 

Electrical Energy. — If a storage battery charged with elec- 
tricity is connected with a motor, work may be done by the 
latter. As the work is done, the electrical condition of the bat- 
tery changes and we therefore ascribe the energy to the battery. 
The energy is called electrical because it is due to a change of 
electrical condition. 

* The amount of heat required to raise the temperature of one pound 
of water one degree Fahrenheit. 



314 WORK AND ENERGY. [Chap. XIV. 

The nature of electrical energy is even less understood than 
that of thermal energy, and no commonly accepted explanation 
of it has vet been made. 



§ III. Principles of Work and Energy. 

306. Principle of Work and Kinetic Energy. — I. For a Par- 
ticle. — The work done in any displacement of a particle by all 
the forces applied to it equals the increment in its kinetic energy 
during that displacement, or 

w = JE & (1) 

if w denote the work done by the forces and JEj. the increment 
in the kinetic energy. The increment is positive or negative 
(and there is a gain or loss of kinetic energy) according as the f?nal 
velocity is greater or less than the initial. 

Proof: Let R denote the resultant of the forces acting on the 
particle, w the work done by them in the displacement, z\ and 
v 2 the velocities of the particle at the beginning and end of the 
displacement respectively. Then, as shown in art. 292, 

iv = I ~R t ds, 

'J S\ 

where 5 is the distance of the moving particle from some fixed 
point in the path and s ± and s 2 are the values of 5 at the begin- 
ning and end of the displacement. Since 

R t = ma t = m dv/dt, 
w = m I 2 ds dv/dt = m I ~v dv = \mv 2 — \mv^. 

Since \mv? and \mv 2 2 are the values of the kinetic energy of 
the particle at the beginning and end of the displacement re- 
spectively, the right-hand member above is the increment in 
the kinetic energy of the particle. 

II. For Any System of Particles. — The work done by all the 
external and the internal forces in any displacement of a system 



§111.] PRINCIPLES OF WORK AND ENERGY. 315 

of particles equals the increment in its kinetic energy during the 
displacement, or 

w # + Wi = JE fc , (2) 

w e and Wi denoting the "external" and the "internal" works 
respectively. 

Proof: According to the principle of work and energy for a 
particle the total work done on a particle in any displacement 
equals the increment in its kinetic energy during that displace- 
ment. Hence the work done on all the particles of a system 
during any displacement equals the sum of the increments in 
their kinetic energies, i.e., the increment in the kinetic energy 
of the system. 

III. For a Rigid Body. — The work done upon a rigid body 
by the external forces in any displacement equals the increment 
in its kinetic energv during that displacement, or 

w, = JE* (3) 

Proof: It is shown in art. 293 that the work done by two 
equal, opposite, and collinear forces is zero for any displacement 
of their application points if the distance between those points 
remains constant. As previously explained, the internal forces 
of any system of particles occur in pairs of equal, opposite, 
and collinear forces, and since in a rigid bod) 7 the distances 
between the application points of the internal forces remain con- 
stant, the work done by the internal forces in any displacement 
of. the body is zero ; hence the work done by the external forces 
equals the increment in the kinetic energy (see II). 

307. Principle of Work and Energy for Conservative Systems. 
— The work done upon a conservative system by external forces 
during any displacement equals the sum of the increments of 
its kinetic and potential energies, or 

w.^JEt + JE,, (1) 

JE p denoting increment of potential energy. 

Proof: Let C x and C 2 be the initial and final and C the 
standard configuration of the system. Also let E p ' and E p " 



3 r6 WORK AND ENERGY. [Chap. XIV. 

denote the potential energies in the configuration C x and C 2 
respectively. The work done by the internal forces in the 
displacement from C x to C would be E p ' ', and that from C 2 to 
C would be E p "\ hence in the actual displacement (C t to C 2 ) 
the work done by the internal forces equals E p ' —E p ". Ac- 
cording to the preceding article, 

w e + (E p '-E p ") = JE k , or w e = JE k + (E p "-E p '). 

Now E p " ' —Ep is the increment in the potential energy of the 
system during the displacement; hence the principle is proved. 

If the work done by the external forces is positive (that done 
by the system against the forces is negative), the system 
gains energy, and if their work is negative (that done by the 
system against them is positive), the system loses energy. 
The statement that a certain amount of positive work, is done 
upon or by a system is equivalent to the statement that the 
system has gained or lost (as the case may be) the same amount 
of energy. 

308. Conservation of Energy. — In any change of condition 
of a material system which is isolated so that it neither receives 
nor gives out energy, its total energy (all forms included) re- 
mains constant in amount; or, as is sometimes stated, "energy 
is indestructible." 

This principle is a generalization based on physical experi- 
ence. It cannot be deduced in the general case from the laws 
of motion, at least not in the present state of knowledge regard- 
ing the constitution of matter and the nature of non-mechanical 
energy. For conservative systems the principle can be proved; 
thus, the system being isolated, there is no external work and 
equation (1), art. 307, becomes 

JE k + JE p = o, (1) 

i.e., the sum of the increments of kinetic and potential energies 
in any change of condition equals zero; hence the sum of the 
kinetic and potential energies is constant. 

If a system not isolated receives or loses energy, some other 
system must lose or receive an equal amount. For, let A be 
the first system and B the one from which A receives or to which 



§111. J PRINCIPLES OF WORK AND ENERGY. 317 

it gives energy. If necessary, imagine B extended so that A 
and B together are an isolated system; then the total energy 
of A and B being constant, if A receives or loses energy, B must 
lose or receive an equal amount. 

309. Principle of Energy for Machines. — The function of a 
machine is to transfer energy from a body or system of bodies 
(A) to another (B). The energy transferred may or may not 
have been transformed in the process. Thus, a dynamo re- 
ceives mechanical and delivers electrical energy, while a water 
motor receives and delivers mechanical energy. 

The energy received by the machine from A is called input 
and that delivered by it to B is called output. The amount of 
energy possessed by the machine at any instant is called its 
stored energy at that instant. It is a fact of experience that 
some of the energy miscarries, as it were, between A and B, and 
is delivered to other bodies than B; that energy is therefore 
called lost energy or simply the loss. This energy is lost prin- 
cipally as heat which is generated wherever there is a transfor- 
mation or transference of energy. 

Let Ei denote the input for any period, 
E the output, 
Ei " loss, 
JE S " increment in the stored energy; 

then, since energy is indestructible, 

E i -(E +E l )=JE SJ 
or . E t -=E.+E x + /E, (1) 

If the stored energy remains constant, or if at the beginning and 
end of the period the stored energies are equal (as at the begin- 
ning and end of a cycle through which the machine works), JE S 
equals zero, and the equation of energy becomes 

E,~E.+E, (2) 

310. Efficiency. — The efficiency of a machine is the ratio of 
the output to input for a period at the beginning and end of 



31 8 WORK AND ENERGY. [Chap. XIV. 

which the stored energy is the same. Thus, if e denotes effi- 
ciency, 

e=E„/E i . (i) 

Since the input is always larger than the output, the efficiency 
of every machine is less than one. 

311. Power or Activity. — The rate at which a machine or any 
"agent" does work is called its power or activity. 

If the work is done uniformly, the power is constant; and if 
Aw denotes the work done in any period At and P the power, 

P = Jw/Jt. (1) 

If the work is not done uniformly, the power is variable and the 
formula above gives the average value of the power ; the actual 
value at any instant is the limit of Aw/ At, or 

p = dw/dt (2) 

Units of Power. — Equations (1) and (2) imply as units of 
power a rate corresponding to unit work done in a unit time. 
Thus one foot-pound-per-second, one meter-kilogram-per-second ; 
one erg-per-second, etc., are such units of power. These units 
are small for some purposes; the following are often more con- 
venient : 

the horse-power =550 foot-pounds-per-second ; 
the force de cheval= 75 meter-kilograms-per-second ; 
the watt = io 7 ergs-per-second.* 



EXAMPLES. 

1. Show that the rate at which steam does work in any 
engine is given by p I an, the notation being: 

p, average steam-pressure per unit area during a stroke; 
/, length of stroke; 
a, area of piston ; 
n, number of strokes per unit time. 

* For dimensions of a unit of power see Appendix C. 



§IV] APPLICATIONS. 3 1 9 

2. Show that the rate at which steam does work in a loco- 
motive whose velocity is v and drivers' diameter is d equals 
2plav/nd. 

§ IV. Applications. 

312. Computation of Velocity and Distance. — The equa- 
tions expressing the principles of work and kinetic energy, (2) 
and (3), art. 306, contain "work terms" on one side and "kinetic 
energy terms" on the other. The factors in the work terms are 
force and distance, and those in the kinetic energy terms are 
mass and velocity. If all the factors except one are known, 
that one may be computed from the equation. In the follow- 
ing examples the unknown quantity is a velocity or a distance, 
and it can be most readily determined by the principle of work 
and kinetic energy. 

EXAMPLES. 

1. Suppose that in ex. 1, art. 290, the velocity of the body 
when at A is 2 ft.-per-sec. What is its velocity when it reaches 
Bl 

Solution: The total work done by all the external forces in 
the motion from A to B was found to be 325 ft. -lbs. (see solu- 
tion of ex. 1). The initial kinetic energy of the body (i.e., at 
A) is 

\mv? = \ 4.97 X2 2 = 9.94 ft. -lbs.; 

hence eq. (3), art. 306, becomes 

325 = \ 4.97^ 2 2 — 9.94, or v 2 = 1 1. 61 ft. /sec. 

2. Suppose that in ex. 2, art. 290, the velocity of the body 
when at A is 5 ft.-per-sec. What is its velocity when it reaches 
Bl 

3. Suppose that in ex. 4, art. 306, the body weighs 10 lbs. 
and that the horizontal surface is rough. What is its velocity 
after having moved 10 ft. ? 

4. Solve the preceding example if the horizontal surface is 
rough, the coefficient of kinetic friction being one-fourth. 

5. How far will the body of ex. 4 move supposing that after 



320 IVORK AND ENERGY. [Chap. XIV. 

the cord slackens the body moves under the influence of fric- 
tion alone ? 

6. Suppose that in ex. 4, art. 266, the angular velocity of 
the rotating body is (jj x at a certain instant. Determine its 
angular velocity when the suspended body has descended a dis- 
tance h after that instant. 

Solution: The external forces acting on the cord and rotat- 
ing and suspended bodies considered as a system are the weights 
of the bodies and the "hinge reaction." Supposing the last to 
be f rictionless , it does no work; neither does the weight of the 
rotating body, but the work done by that of the suspended 
body is Wh. The work done by forces internal to the two 
rigid bodies and that by the reactions between the cord and 
the two bodies is zero. The work of the forces internal to the 
cord is not zero, but it is small (see art. 315) and is negligible in 
this instance. 

The initial kinetic energy of the system (neglecting the mass 
of the cord) is 

\Iu) 2 +\m{ru)^ 2 , 

and if co 2 denotes the final angular velocity, eq. (2) art. 306 
becomes, since a>i = o, 

Wh=^Iaj 2 2 + ^m{roj 2 ) 2 -^Iw^^-^m(roj x ) 2 , 

from which co 2 can be computed. 

7. Suppose that the disk of ex. 2, art. 298,1s rotating on a 
shaft 4 in. in diameter and that the axle friction is 30 lbs. In 
how many turns would the frictional resistance stop it ? 

8. Let h denote the height of the centre of gravity of a pen- 
dulum when the velocity is zero above its lowest position. Show 
that the angular velocity of the pendulum when it reaches its 
lowest position is (i/k)x / 2gh, k denoting the radius of gyration 
of the pendulum with respect to the axis of suspension. 

9. A body is suspended by two parallel cords of equal length 
and is allowed to swing in the plane of the cords under the in- 
fluence of gravity. Show that the height (h) to which the cen- 
tre of gravity rises above its lowest position and the velocity (v) 
in the lowest position are related as follows : v 2 = 2gh. 



§IV.] APPLICATIONS. 321 

313. Train Resistance. — The resistance to motion experi- 
enced by a train moving on a track consists of rolling resistance, 
journal friction (at the car and locomotive axles), air resistance, 
and the frictional resistance at the working parts of the loco- 
motive. For simplicity we may imagine these replaced by a 
single horizontal resistance equivalent to them all so far as the 
motion of the train is concerned. This single equivalent resist- 
ance is often called train resistance, but sometimes the term is 
intended not to include the locomotive and tender resistance. 

It is customary to express train resistance as so many pounds 
for each ton of weight of train. Many experiments have been 
made to determine train resistance, and a number of formu- 
las have been deduced to express its value. We give but three 
as illustrations and to furnish data for a few examples; the 
notation is as follows: 

R, train resistance in lbs.-per-ton; 

V, velocity in mi.-per-hour; 

T, weight of train in tons (2000 lbs.). 

Engineering News : R = V / \ + 2 (1) 

This includes all resistances except the internal friction 
of the locomotive. For the higher velocities it was deduced 
from experiments on a fast passenger- train. 

Crawford: R = V*/26S + 2.s (2) 

This does not include resistance of locomotive and tender. 
It was deduced from experiments on passenger-trains. 

Lundie: R = 4 + F[o.2 + 14/(35 +T)] (3) 

This was deduced from experiments on electric elevated 
street-railway trains and includes all resistances. 
It will be noticed that only (3) contains T. The first two 

apply only to trains of approximately the same kind and, 

weight as those experimented on. 



322 WORK AND ENERGY. [Chap. XIV. 

EXAMPLES. 

i. Plot the curves represented by the preceding equations, 
assuming T in the third to be one ton. 

2. What is the relation between the "draw-bar pull" be- 
tween tender and first car when the velocity of the train is con- 
stant ? 

Solution: The only forces doing work on the train are the 
draw-bar pull and the train resistance. Since the velocity of 
the train is constant, its kinetic energy remains constant and 
the total work done by the two forces must be equal to zero 
(see eq. (2), art. 306), i.e., the works done by them must be 
equal but of opposite sign. Since the forces act through equal 
distances in any motion of the train, the forces must be equal 
in amount. 

Train resistance (for the cars) is determined by measuring 
the draw-bar pull in front of the first car when the velocity is 
constant. 

3. Assume the train resistance for cars to be constant (as it 
is roughly below 10 mi.-per-hr., according to Crawford's for- 
mula) and that it equals 2.6 lbs.-per-ton. If the cars with load 
weigh 1000 tons, how much work does the engine do upon the 
cars to bring up the velocity of the train from o to 10 mi.-per-hr., 
if it is done in a distance of one mile on a level track ? 

4. Solve the preceding example supposing that the train 
runs upa" one-per-cent grade." * 

5. Supposing that the speed in ex. 3 is increased uni- 
formly, i.e., the acceleration is constant, what is the value of 
the draw-bar pull? 

6. Express the rate at which the engine does work on the 
train in the preceding example in horse-powers when the train 
is starting and when its velocity is 10 mi.-per-hr. 

7. Show that if P denotes the rate at which work is done by 
the steam in a locomotive, i.e., the rate at which energy is sup- 
plied to the engine, R the total train resistance, and v the speed, 
and if v is constant, P=Rv. 

* One ft. rise for every 100 ft. along the track. 



§ I V. ] AP PLICA TIONS. 323 

Solution: The speed being constant, there is no change in 
the kinetic energy of the train ; therefore the work done on the 
engine by the steam equals the work done by the train against 
the resistance during any period. If w denotes the first work 
during which the train moves any distance s, in a time t, 

w = Rs, or w/t = Rs/t. 

Now w/t is the rate at which work is done on the engine, and 
s/t is the velocity of the train; hence P = Rv. 

8. If the Engineering News formula is correct at all speeds 
and gives practically the whole train resistance, show that the 
cylinder steam-pressure per unit area (average for one stroke) 
required to maintain any constant speed in a train on a level 
track is a linear function of the speed. (See ex. 2, art. 311.) 

9. Show that when the speed of a train is changing, P = 
Rv + mva, m and a denoting the mass and acceleration of the 
train respectively and P, R, and v having meanings as in ex. 7. 
(The equation neglects the "rotational" component of the 
kinetic energy of the wheels, which is small compared to the 
kinetic energy of the train.) 

314. Friction Brakes. — Fig. 235 represents a form of friction 

brake often used to measure the power of ^ ^ 

small motors. It consists essentially of a flat- f \ 

faced pulley rigidly fastened to the shaft of [ c «___ r __ 
the motor, a strap or rope partly encircling the j\ J 

pulley, one end of it being fastened to a 1 \^__^/ I 
spring-balance and the other sustaining a freely HJ [J 

hanging body. v±- 

When the pulley rotates it drags the strap Fig. 235. 

around with it a small distance and the spring tension (T) is 
greater or less than W according as the rotation is clockwise or 
counter-clockwise. Assuming it to be clockwise, the frictional 
resistance at the rim of the wheel equals T — W and the work 
done by the motor against friction in one revolution of the 
wheel is (T — W)2nr. Hence, if the wheel makes n revolutions 
per unit time and all of the work done by the motor is thus 



3^4 



WORK AND ENERGY. 



[Chap. XIV. 



expended against friction, the power of the motor equals 

(T — W)27trn. 

Fig. 236 represents a form of brake sometimes used on hoist- 
ing-drums for stopping the same. 
It consists essentially of a strap 
partially encircling the drum or a 
wheel fastened to the drum, one 
end being fastened to a fixed sup- 
port, as A, and the other end to a 
lever,, as CD. A force (P) applied 
as shown brings a frictional resist- 
ance to bear upon the drum at the 
strap and so controls the speed. We 
wish to find the relation between 

the pressure P and the weight of the descending load when its 

velocity is constant. 

Let T 1 and T 2 denote the tensions in the strap at A and B 

respectively, F the frictional resistance at the brake-strap, / the 

coefficient of friction, F' the axle friction, and f the coefficient 

of axle friction. Since the kinetic energy of the moving system 

is constant, 




Wi 



F27Za+F'27lC (1) 



As shown in art. 140, 

7>7>K (2) 

Considering the forces on the lever, it is seen that 



T 2 d = Pb. 



Also, 
and 



F' = f'(W+7\ + T 2 ) . 
F=T 2 -T l 



(3) 
(4) 
(5) 



These five equations determine all the unknown quantities P, 
T lt T 2 , F, and F'. 

315. Efficiency of Tackle. — Fig. 237 represents a fixed pul- 
ley. Let a denote its radius, r that of the axle, and d the diam- 
eter of the rope ; also let P and Q denote the tensions on the two 
sides as shown. 



§IV.] 



APPLICATIONS. 



3 2 5 



Consider the pulle) r and as much of the rope as is shown 
together as a machine. In any motion, the external works are 
those done by P, Q, and the axle friction; the internal work is 
done by friction between the rope fibres where the rope winds 
on and off. In one revolution of the wheel the works done by 
P and Q respectively are P2-C1 and —Q27za, 
and that done by the axle friction is— f'R.27ir, 
f denoting the coefficient of axle friction and 
R the resultant axle pressure. Considering 
that numerical values of f are uncertain, we 
may write R = 2Q; then the work of axle 
friction is approximately —f'Q^nr. It has 
been found experimentally that the work 
"due to rigidity" of the rope is proportional 
to Q, i.e., to the tension on the following 
side; hence we may write for this work in 
one revolution —cQ27rr, c being an experimental coefficient. 

The equation of work and energy is (for one revolution) 




Fig. 237, 



or 



P27za — Q2na — f'QAxr — Q27ZC = o, 
P = Q(i+c + 2f'r/a)=Qk, . . 



CO 



k being an abbreviation for (1 +c + 2fr/a) and sometimes called 
1 ' coefficient of resistance ' ' for the pulley. 

With equation (1) we can compute the relation between the 
load and the necessary pull to raise or lower it, and the efficiency 
of the tackle (see ex. 1). The value of the coefficient c has been 
found to vary directly as the square of the diameter of the rope 
(d) and inversely as the radius of the pulley. One formula 
(Eytelwein's) is 

c = o.46d 2 /a, 



d and a being expressed in inches. If a = 4<i, r — d/2, and 
/' = o.i, the values of k are as follows: 



d = 


% 


f 


I 


l£ 


k = 


1.08 


I . II 


1. 14 


1 . 20 



326 WORK AND ENERGY. [Chap. XIV. 

EXAMPLES. 

i. Determine the relation between F and W (fig. 8$) and 
the efficiency when the load is being raised. 

Solution : Call the tensions in the sections made by the hori- 
zontal line beginning at the left 5 lf 5 2 , 5 3 , etc. Then S 2 = kS v 
S 3 = k 2 S lt S 4 = k 3 S v etc., and since 

W=S 1 + S 2 +S 3 + etc., 

Also F = kS 6 = k Q S 1 ; 

hence F = Wk 6 /(i +k+k 2 + k 3 + k* + k*). 

During an ascent of the load equal to s, the point qf appli- 
cation of F descends a distance 6s. Hence for that motion the 
output and input are respectively 

Ws and 6Fs, 
ano the efficiency equals 

W/6F = (i+k + k 2 + . . .k 5 )/6k\ 

2. Solve the preceding example supposing that the load is 
being lowered. 

316. Efficiency of a Mine-hoist. — Fig. 238(a) represents a 
balanced vertical mine-hoist, consisting of an endless cable, 
two cages, a hoisting-drum above, and a pulley below. Ordinates 
from Ot to the various lines in fig. 238(6) represent various 
quantities involved in a power computation which is now to be 
made. Thus, ordinates to aa represent a frictional resistance 
(2 tons) applied to the surface of the drum equivalent to all the 
actual frictional resistances in the hoist; ordinates to bb repre- 
sent the weight of the load (6 tons) ; and ordinates to cc repre- 
sent the velocities of the cages. The assumed law of motion is 
(1) that the acceleration is greatest at the beginning of the 



§ iv.} 



APPLICATIONS. 



2> 2 1 



motion and decreases to zero uniformly and so that the velocity 
acquired in 30 sees, is 60 ft.-per-sec, (2) then the velocity re- 
mains constant for 60 sees., (3) then a retardation follows which 
increases just as the acceleration decreased, i.e., uniformly and 
so that the cages are brought to rest in 30 sees. 

When the speed of the cages is constant the pull (P) of the 
hoisting-engine (assumed for simplicity as applied to the sur- 
face of the drum) just equals the sum of the load and the 



□ 








100lt.i>ursee. 



(a) 



(b) 




Fig. 238. 



friction, but in the first 30 sees, the pull is increased by the 
" inertia force," and in the last 30 sees, it is decreased by 
the "inertia force." This inertia force depends upon the 
mass of the cable, load, and cages, the moment of inertia 
of the drum, and the acceleration or retardation. We 
suppose that these are such that the ordinates to dd' and 
to d"d represent the values of the forces as applied to the 
surface of the drum and that its maximum value is 12 tons. 
The total force then to be exerted by the engine on the drum at 
any instant is represented by the sum of the- ordinates to the 
three lines aa, bb, and dd for that instant. Ordinates to the 
line ee represent all such sums. The work done by the engine 

in hoisting the load a distance 5 from the bottom equals / Pds, 

and the rate at which the engine works (i.e., its power) at the 
instant when the velocity of the cages is v equals Pv. 



328 WORK AND ENERGY. |C HAP - XIV - 



EXAMPLES. 

i. Draw a curve showing how the power changes with the 
time. 

2. Determine the entire work done by the engine in a single 
hoist, and also the efficiency of the hoist. 



CHAPTER XV. 

IMPULSE AND MOMENTUM. 

§ I. Impulse. 

317. Impulse of a Force whose Direction is Constant. — If 

the magnitude of the force is constant, the impulse of the force 
for any interval is the product of the force and the length of the 
interval, i.e., if F and (t" — f) denote the force and the interval 
respectively, the impulse equals F(t" — t'). 

If the magnitude of the force varies, the impulse for any ele- 
ment of time equals the product of the value of the force at any 
instant of the element and the length of the element, i.e., Fdt; 

and the impulse for any finite interval t" — t' equals / Fdt. 

The unit of impulse depends on the units used to express force 
and time. If C.G.S. units be used, the unit of impulse is called 
a dyne-second; if the pound and second are used as units of force 
and time respectively, the unit of impulse is called a pound- 
second.* 

318. Impulse of a Force whose Direction Varies. — An impulse 
should be regarded as a vector quantity, its direction being the 
same as that of the force if that is constant. 

The meaning of impulse of a force whose direction varies may 
be explained as follows : Imagine a force to change its direction 
and magnitude five times in a certain interval, and that the 
values of the force and the corresponding portions of the inter- 
val are F' , F", etc., and (At)' , (At)" , etc., respectively. Then 
the impulses of the force for these portions are F'(Jt)', F"(4t)", 
etc. Now if Aa, ab, etc. (fig. 239), represent these impulses, the 
impulse of the force for the entire interval is the vector sum of 
Aa, ab, etc., or AB. 

* For dimensions of a unit impulse see Appendix C. 

329 



33° 



IMPULSE AND MOMENTUM. 



[Chap. XV. 




Fig. 239. 



If the force changes by small amounts and many times in 
the interval, its variation may resemble that of a continuously 
varying force. By impulse of a continuously varying force is 
meant the limit toward which the impulse of a suddenly varying 

force tends as the manner of varia- 
tion of the latter approaches that of 
the former. 

Employing the language of the 
infinitesimal calculus again, we state 
that the impulse of a force for an ele- 
ment of time equals F-dt if F de- 
notes the value of the force at any 
instant of the element, and its direc- 
tion is that of the force. The im- 
pulse of the force for a finite interval is / Fdt, the integration 

being not ordinary but vectorial.* 

319. Component of an Impulse. — Since an impulse is a vector 
quantity it can be resolved; thus let AB (fig. 239) represent an 
impulse and OX an x axis ; then the x component of the im- 
pulse is represented by AC. 

Proposition. — The component of the impulse of a force along 
any line equals the impulse of the component of the force along 
that line. 

Proof : Let F denote the value of the force at any instant and 
a its angle with the line (the x axis, say). Then for any element 
of time (dt) including the instant, the x component of the im- 
pulse of the force is (Fdt) cos a. Evidently the x component of 

the impulse for a finite interval t" — f equals the sum of the x 

Xt" 
Fdt -cos a. But 



f 



Fdt -cos a 



f F * 



dt, 



F x denoting the x component of F. Since the second integral 

* While the student may not be able to compute the impulse of a 
force in this way, it is desirable that he should understand the principle 
of the method as above given. 




§ I.] IMPULSE. 33 1 

is the impulse of the x component of the force for the interval f 
the proposition is proved. 

320. Moment of the Impulse of a Force. — We regard the im- 
pulse of a force as having not only magnitude and direction, but 
also "position." If the action line of the force is fixed, then 
that line is also the position line of the impulse. If the action 
line changes, then the position line of the impulse for an element 
of time coincides with the action line of the force at any instant 
of the interval. 

I. The Force is Constant and its Action Line is Fixed. — Let 
ab (fig. 240) represent such a force (F) and OX an axis of mo- 
ments. The impulse of the force for a 
period t" — t' is F{t" — t'), its position line 
being ab. 

The moment of the force about OX is 
defined in art. 28 as the product of the 
component of the force which is perpen- 

dicular to OX (the other being parallel to ~™ 

Fig. 240. 
OX) and the distance between the perpen- 
dicular component and the axis. That is, if M denotes the 
moment and p the distance, 

M=(Fsin a)p (1) 

In an analogous way we define the moment of the impulse of 
the force (or the angular impulse of the force, as it is also called) 
to be the product of that component of the impulse which is 
perpendicular to the axis and the distance between that com- 
ponent and the axis. If ab is taken to represent the impulse, 
ac represents the perpendicular component, and the angular im- 
pulse equals 

F(t"-t')sma.p=M(t"-t'), .... (2) 

i.e., for any period the moment of the impulse about any axis 
of a constant force whose action line is fixed equals the product 
of the moment of the force about that line and the length of 
the period. 

II. The Force Varies in Any Way. — The moment of the im- 
pulse about any line for an element of time is the product of the 



33 2 IMPULSE AND MOMENTUM. [Chap. XV. 

value of the moment of the force about that line at any instant 
of the element and the element, i.e., Mdt. The moment of the 

impulse for a finite interval t" — t' is the sum of the angular im- 

Xt" 
Mdt. 

The Unit of an Angular Impulse is the angular impulse 
of a force whose impulse is such that its component perpendic- 
ular to a moment axis equals a unit impulse and has an arm of 
unit length. There are no names in use for these units. To 
describe the unit of any numerical value of an angular impulse, 
or moment of an impulse, we name the units of impulse and 
length used in the computation. 

The rule of signs for moments of impulses is like that for 
moments of forces (see art. 28), i.e., we give the same sign to the 
moment of an impulse of a force with respect to an axis as we 
give to the moment of the force with respect to that axis. 

§ II. Momentum. 

321. Momentum of a Particle. — The momentum of a particle 
is the product of the mass (m) and the velocity (v) of a particle. 

Unit of Momentum. — The definition implies as unit the mo- 
mentum of a particle of unit mass moving with unit velocity. 
The magnitude of the unit hence depends upon the units of mass 
and velocity employed. No single words have been generally 
accepted as names for any units of momentum. It is shown in 
Appendix C that the dimensions of a unit momentum are the 
same as those of a unit impulse; hence it is not inappropriate 
to call these units by the same name. Thus in the C.G.S. sys- 
tem the unit of momentum is called a dyne-second, in the 
English engineers' system it is called a pound (force)-second, etc. 

322. Components of a Momentum. — Momentum should be 
regarded as a vector quantity, its direction being the same as 
that of the velocity of the particle. Like any other vector quan- 
tity, a momentum may be resolved; thus if od (fig. 241a) rep- 
resents the velocity of a particle, to some scale it also represents 
the momentum {mv), and oa and oa' represent two components 
of the momentum , their values being mv cos a and mv sin a 



II.] 



MOMENTUM. 



333 



respectively. Also oa, ob, and oc are the x, y, and z compo- 
nents of the momentum, and their values are 

mv cos a = mv x , mv cos /? = mv^, mv cos y = mv z . 

323. Moment of Momentum. — Momentum should be regarded 
as having not only magnitude and direction, but also "position/ 





(a) 



C6) 



Fig. 241. 



The tangent line along which the velocity at any instant is 
directed is the position line of the momentum. 

Analogous to the moment of a force with respect to an axis 
(art. 28) we define the moment of momentum of a particle (or 
angular momentum, as it is also called) to be the product of that 
component of the momentum which is perpendicular to the axis 
(the other being parallel to it) and the distance between the per- 
pendicular component and the axis. Thus, the momentum rep- 
resented by od (fig. 241a) has a moment about the line OX equal 
to oa' Xp, p denoting the distance from the axis to oa' . 

The unit of angular momentum is the angular momentum 
of a particle whose momentum is such that its perpendicu- 
lar component equals a unit momentum and has a unit " arm." 
There are no names in use for these units. To describe the unit 
of any numerical value of a moment of momentum we name the 
units of momentum and length used in the computation. 

The rule of signs for moments of a momentum is similar to 
that for moments of forces (art. 28). We imagine the "perpen- 
dicular component" to be a force and then the sign of the 



334 IMPULSE AND MOMENTUM. [Chap. XV; 

moment of momentum is like that of the moment of that force. 
Thus, in the preceding illustration, the angular momentum 
about the x axis is positive. 

Proposition. — If the momentum of a particle be resolved into 
three rectangular components, the moment of momentum with 
respect to a line parallel to one of the components equals the 
sum of the moments of the other two components with respect 
to that line. (Compare Prop. I, art. 28.) 

Proof: Let od (fig. 241) represent the momentum, OX, OK, 
and OZ directions of resolution, and OX the moment axis. As 
explained in the foregoing, the moment of momentum with 
respect to that axis is 

oa'Xp = (mv sin a)p. 

From fig. 241(6), which represents the lines in the plane hoc in 
their true relations, it is plain that 

p = y cos f — z sin 7'; 

hence the moment of momentum equals 

mv sin a (y cos y f — z sin y') y 

or (mv z )y — (mv y )z. 

324. Momentum of a System of Particles. — By momentum of 
a system of particles is meant the resultant * of the momenta 
of the particles. We shall not need a general expression for this 
resultant, but will deduce the value of its component along any 
line and the value of the resultant in special cases (art. 326). 

Just as in a system of forces, the component of the resultant 
of any number of momenta along a line equals the algebraic 
suni'of the components of the momenta along that line. Thus 
the x component of the resultant equals, if m' , m" ', etc., denote 
the masses of the particles and v', v" , etc., their velocities, 

m'v x '+m"v x "+ . . . =Imv x =Mv x (see eq. (2), art. 239), 

* Computed according to the methods for compounding forces as 
given in Chap. II. 



§ II.] MOMENTUM. 335 

M being the mass of the system and v the velocity of its mass- 
centre. That is, the component of the momentum of a system 
of particles along any line is the same as if the entire mass were 
concentrated at the mass-centre. 

325. Moment of the Momentum of a System of Particles. — 
By moment of momentum of any system of particles about any 
line is meant the algebraic sum of the moments of the momenta 
of the particles about that line. 

As explained in art. 323, the moment of the momentum of a 
particle whose mass, velocity, and coordinates are ni' , v' t 
{%' , y f , z') about the x axis is 

{m'v z 'y' — m'VyZ) ; 

hence the moment of momentum of the system about the x axis 
is 

iXmVzy — mVyZ). 

326. Momentum of a Rigid Body in Special Cases. — I. A 

Translating Body. — The velocities of all particles being the same 
in magnitude and direction, the momentum equals, if v denotes 
the common velocity, (dm) lt (dm) 2 , etc., the masses of the par- 
ticles, and m the mass of the body, 

(dm^v + (dm) 2 v + ■ • ■ =vi'dm = mv, 

and the direction of the momentum is the same as that of the 
velocity. 

The position line of the momentum contains the mass-centre, 
as can be shown by a method similar to that employed in art. 
244. Hence the moment of the 
momentum about any line is the 
same as if the entire mass were con- 
centrated at the mass-centre. 

II. A Rotating Body. — We assume 
as in art. 262 that the rotating body 
is homogeneous and has a plane of 
symmetry perpendicular to the axis 
of rotation. Let fig. 242 represent 

that section of symmetry, C the mass-centre, and the centre of 
rotation. As in art. 242, imagine the body divided into ele- 




33& IMPULSE AND MOMENTUM. [Chap. XV. 

mentary rods parallel to the axis of rotation. Evidently the 
position line of the momentum of each rod is in the plane of 
symmetry ; hence that of the momentum of all the rods (or the 
body) must be in that plane. Since the components of the 
momentum of the body parallel to the x and y axis equal mv x 
and mvy respectively (see art. 324), the resultant momentum 
equals 

[(mv x ) 2 + (mv y ) 2 ] =n, =mr(o 

(oj denoting the angular velocity of the body), and its direction 
is the same as that of the velocity of the mass-centre. The posi- 
tion line of the momentum passes through a point Q in the line 
OC, whose distance (q) from the axis of rotation is given by 

q = k 2 /?, 

k being the radius of gyration of the body with respect to the 
axis of rotation. This fact can readily be proved from the 
result of the remainder of this article. 

The moment of the momentum about the axis of rotation 
might be computed from art. 325, but the following method is 
as simple and preferable: Let P(fig. 242) represent any particle, 
dm its mass, and r its distance from the axis of rotation; then 
the momentum of that particle is dmrco, its position line as 
shown, and the moment of its momentum equals (dmrco)r. 
Hence the moment of the momentum of the body equals 



/ dmr 2 (D = co I dm'T 2 =Iaj i 



I denoting the moment of inertia of the body with respect to 
the axis of rotation. This resultant is independent of the as- 
sumption made in the first paragraph. 

§ III. Principles of Impulse and Momentum. 

327. Principles for a Particle. — Let a, v, and (x, y, z) denote 
the acceleration, velocity, and coordinates at any instant of a 
particle whose mass is m, and let R denote the resultant of all 
the forces applied to it. Then, according to art. 236, 

R x = ma x = mdv x /dt, 
or R T dt = mdv x . 



§ III.] PRINCIPLES OF IMPULSE AND MOMENTUM. 337 

Let v' and v" denote values of the velocity at times t' and t'\ 
respectively, then 

jf/R x dt = mv x "-mv a / (i) 

The integral is the x component of the impulse of the resultant 
for the interval {t" — t'),and the right-hand member is the incre- 
ment in the momentum of the particle along the x axis during 
that interval. Now the component along any line of the im- 
pulse of the resultant of any number of forces applied to a par- 
ticle equals the algebraic sum of the components of the impulses 
of the forces along the same line ;* hence 

The algebraic sum of the components along any 
line of the impulses of the forces applied to a 
particle equals the increment in the component of 
the momentum of the particle along that line. 
From art. 236 we have also 

Ry = mdv y /dt and R z —mdv z /dt\ 
hence R y z = mzdv y /dt, R z y = mydv z /dt, 

and (R z y — R y z) = mydv z — mzdv y . 

Now the left-hand member is the moment of the resultant about 
the x axis, and we will replace it by M x . The left-hand member 
equals d(mv z y — mv y z)/dt, i.e., the time-rate of increase of the 
moment of momentum about the x axis (see art. 323), and for 
convenience we will replace it by U. Then 



and 



M x = dU/dt, or M x dt = dU, 

H'' 



j M,dt = U"-U'. (2) 



The left-hand member equals the moment of the impulse of the 
resultant about the x axis (equals also the sum of the moments 
of the impulses of the forces acting on the particle), and the 
right-hand member is the increment in the moment of the mo- 
mentum of the particle about the x axis. Hence 

"* The proof can be supplied by the reader (see art. 319) 



33 8 IMPULSE AND MOMENTUM. [Chap* XV. 

The algebraic sum of the moments of the impulses 
of the forces applied to a particle about any line 
equals the increment in the moment of the mo- 
mentum of the particle about that line. 

328. Principles for a System of Particles. — The internal forces 
in a system of particles occur in pairs, the forces of each being 
at each instant equal, opposite, and collinear. Therefore the 
impulses of each pair of forces for any interval are equal and 
opposite; also the moments of the impulses of each pair about 
any line are equal and opposite. It follows that the internal 
forces contribute nothing to any change in the momentum or 
moment of momentum of a system of particles, and it can be 
readily shown from the results reached regarding a single par- 
ticle that 

A. For any period the algebraic sum of the components of 
the impulses of the external forces acting on a system along any 
line equals the increment in the component of the momentum 
of the system along that line. 

B. For any period the sum of the moments of the impulses 
of the external forces acting on any system about any line 
equals the increment in the moment of the momentum about 
that line. 

Special Case: No External Forces Applied to the System. — 
It follows from the foregoing that 

(a) The component of the momentum of the system along 
any line remains constant; this principle is called that of "con- 
servation of linear momentum." 

(b) The moment of the momentum of the system with re- 
spect to any line remains constant; this principle is called that 
of "conservation of angular momentum." 

§ IV. Applications. 

329. Computation of Velocity and Time. — The equations 
expressing the principles of impulse and momentum (A and B } 
art. 328) contain impulse (or moment of impulse) terms on one 
side and momentum (or moment of momentum) terms on the 
other. The factors in the impulse or moment of impulse terms 



§ IV.] APPLICATIONS. 339 

are time and force or moment of force, and those in the momen- 
tum or moment of momentum terms are velocity and mass or 
moment of mass. If all the factors except one in one of the 
equations are known, that one may be computed. In the fol- 
lowing examples the unknown quantity is a velocity or a time, 
and the»examples can be most readily solved by the principles of 
impulse and momentum. 

EXAMPLES. 

i. A body weighing 80 lbs. is moved along a horizontal sur- 
face by a horizontal pull of 100 lbs. If the frictional resistance 
is 20 lbs., how much velocity does the body acquire in 10 seconds ? 

Solution: There are three external forces acting upon the 
body, the pull, the weight, and the reaction of the surface. The 
components of the impulses of these along the path are respect- 
ively 

100X10, o, and — 20 X 10 lb. -sees. 

Calling the velocity of the body at the beginning of the 10-sec. 
period v v and that at the end of it v 2 , the equation of impulse 
and momentum for the period is 

1000 — 200 = 2.45^3 — 2.45^, 

2.45 being the mass of the body in geepounds. Hence the in- 
crement of the velocity is 

i' 2 — z\ = 800/2. 4 5 = 326.5 ft.-per-sec. 

2 . Solve ex. 1 , supposing that i\ = o , and that the pull instead 
of being constant equals 20 + io£ (t being time in sees, after the 
instant of starting). 

3. Supposing that v 2 in ex. 1 is 300 ft.-per-sec, determine 
how long the body would slide after the tenth second, the 
motion taking place under the influence of friction alone. 

4. Suppose that the disk of ex. 2, art. 298, is rotating on a 
shaft 4 in. in diameter, and that the axle friction is 30 lbs. In 
how many seconds would the frictional resistance stop it ? (Use 
B, art. 328.) 

5. Two pulleys are mounted on the same shaft, one being 



340 IMPULSE AND MOMENTUM. [Chap. XV. 

" fast " and the other " loose." Suppose that the shaft (and the 
fast pulley) to be turning at a certain instant with an angular 
velocity co, and that the loose pulley is quickly made fast so that 
it also turns with the shaft. If the shaft turns in smooth bear- 
ings, determine the subsequent angular velocity of the pulleys 
and shaft. 

Solution : Under the supposition there are no external forces 
acting on the rotating system having moments about the axis 
of the shaft. Then the moment of momentum of the three 
bodies about the axis of the shaft remains constant. 

Let I denote the moment of inertia of the shaft and the fast 
pulley, V that of the loose pulley, and co 2 the final velocity. Be- 
fore and after the loose pulley is made fast, the moments of 
momentum of the three bodies about the axis are respectively 

Iaji + o and Ico 2 + I'co 2 ; 
and since these are equal as above explained, 

co 2 = IcoJ{I + F). 

6. Two spheres whose masses and radii equal m and r respect- 
ively rotate on a light frame about a vertical axis as shown in 
fig. 243 with an angular velocity w v Suppose that, in some way 



K- — 



en 



W 

Fig. 243. 

without interfering directly with the motion, the distances a 
are increased to b. Determine the angular velocity after the 
change. 

330. Pressures Due to Jets. — Pressure due to a jet of liquid 
can often be most readily determined by the principles of im- 
pulse and momentum. How this is done is explained in the 
solution of some of the following 

EXAMPLES. 

1. Fig. 244(a) represents a jet impinging on a flat surface so 
that the direction of the jet is changed 90 . If W is the weight 



IV.] 



APPLICATIONS. 



341 



(of water) impinging per unit time and v the impinging velocity, 
show that the pressure of the water on the plate is Wv/g. 

Solution: Consider the motion of the amount of water rep- 
resented in the figure for a small period dt, during which it moves 




(a) 



Fig. 244. 



(0) 



into the position indicated by the dotted lines. Let M x and 
M x " denote the initial and final momentums in the x direction 
of the water whose motion is being considered, then 

M x ' =the "x momentum" of B + (Wdt/g)v; 
also Ma/' = the "x momentum" of B, 

because C and D have no x momentum. Hence the change in 
the x momentum of the body of water is 

, M x '-M x " = (Wdt/g)v. 

The impulse of the force producing this change (the pressure of 
the surface on the water, which call F) is Fdt; hence 

Fdt = (Wdt/g)v, 

or F = Wv/g. 

2. Fig. 244(6) represents a jet impinging on an inclined sur- 
face. If W is the weight of the water impinging per unit time, 
and v the impinging velocity, show that the normal pressure on 
the jet equals (Wv/g) cos <j>. 

Solution: Consider the motion of the body of water repre- 
sented for a small period during which it moves into the position 
indicated by the dotted lines. With notation as in the preced- 
ing solution, 

M x =x momentum of B + (Wdt/g)v cos <j) 
and M x n = x momentum of B. 



342 



IMPULSE AND MOMENTUM. 



[Chap. X\ 



oo' 






oo 



Hence the change in the x momentum of the body of water is 

MJ-MJ' = (Wdt/g)v cos 4>. 

Calling the normal pressure of the surface on the water F n , the 
impulse of F n is F n dt, and 

F n dt = (Wdt/g)v cos <£, or F n = (Wv/g) cos <f>. 

3. Fig. 244(c) represents a jet impinging on a guide or vane 
which suppose smooth, so that the speed of the water is not 
changed. Determine the x and y components of the pressure 
of the jet. Ans. The x component is Wv(i —cos <j>)/g. 

4. Fig. 245 represents in plan and elevation a simple reaction- 
wheel. Water is poured in at the top 
and escapes through two orifices O and 
O in a horizontal direction. Such a 
wheel is caused to revolve by the "re- 
action" of the jets. It is required to 
determine the moment of these reactions. 

Solution: Let W denote the weight 
of the water escaping and entering per 
unit time, and v the velocity of the 

|c /5?^\ escaping water relative* to the orifices. 

Then if co denotes the angular velocity 
of the wheel, the absolute velocity of the 
jets is rco—v. Consider the motion, for 
Fig. 245. a short period, of the water in the wheel 

and the amount about to flow in for that period (Wdt); that 
water is represented in the elevation fig. 245. At the end of 
the period the entering water is all in the wheel and an amount 
Wdt has escaped, as shown in the plan. Let M f and M" denote 
the moment of momentum of the body of water being consid- 
ered about the axis at. the beginning and end of the period di 
Then supposing that the water enters vertically or so that th 
jet is bisected by the axis of rotation, 

M' = the moment of momentum of B 
and M"= " " " " "B + (Wdt/g)(r(o-v)r; 

the change of moment of momentum hence equals 
(Wdt/g){rco-v)r. 



ELEVATION 




§ IV.] APPLICATIONS. 343 

Let M a denote the moment (about the axis) of the pressure of 
the wheel on the water; then the moment of the impulse of 
the pressure is M a dt and 

M a dt = (Wdt/g)(rco-v)r, 
or M a = (W/g)(raj-v)r, 

and the moment of the water-pressure (turning the wheel) is 

(W/g)(v — raj)r. 

5 . Show that when water issues from an orif.ce in a vessel at 
rest, the water exerts on the vessel a force equal to (W/g)v 
in a direction opposite to that of the velocity of the jet. (W de- 
notes the weight of water escaping per unit time, and v its 
velocity.) 

331. Sudden Impulses. — The impulse of a force which acts 
for a very short time is called a sudden impulse. 

If a body is subjected to a blow and to an ordinary or steady 
force at the same time , the impulse of the latter during the blow 
is often negligible compared with the sudden impulse. Thus in 
the case of a ball thrown against a wall there are two forces 
acting on the former during the impact, namely, the weight of 
the ball and the reaction of the wall. Supposing the ball to be 
thrown horizontally, the reaction of the wall is horizontal, and 
its impulse equals the change in the horizontal component of 
the momentum of the ball. The impulse of the weight equals 
the change in the vertical component of the momentum. Now 
this latter change, as we see from observation, is practically zero 
compared to the first change; hence the impulse of the weight 
is also practically zero compared to the impulse of the blow. 

The principles of impulse and momentum are especially 
adapted to questions involving sudden changes of motion, and 
the remainder of this chapter relates to changes of this kind. 

332. Force of a Blow and Recoil of a Gun. — When one body 
strikes another we name the act "a blow," and by force of the 
blow we mean the pressure between the bodies during the blow. 
This pressure is variable, changing from zero to a maximum and 



344 IMPULSE AND MOMENTUM. [Chap. XV. 

back to zero. If t denotes the duration of the blow andF the 

variable value of the force, the impulse is / Fdt. By average 

force of the blow * is meant a constant force which, acting for 
a time equal to the duration of the blow, has an impulse equal 
to that of the actual force. Thus if F a denotes the average 
force of the blow, 

F a t = f o Fdt. 

When a gun is fired, the powder-gases exert a backward force 
on the gun as well as a forward one on the shot. If the mass of 
the powder were negligible, the two forces at each instant and 
their impulses would be equal. Thus let m 1 and m 2 denote 
masses of the shot and gun respectively, and v 1 and v 2 their 
velocities just after the shot leaves the barrel. Then, since the 
impulses producing the velocities are equal, the momenta of 
shot and gun must be equal, i.e., 

m 1 v 1 = m 2 v 2 . 

If a gun is suspended in a horizontal position by means of 
two parallel cords of equal length, the velocity of recoil (v 2 ) and 
the height (h) to which the gun rises during the recoil are related 
thus: v 2 2 = 2gh (see ex. 9, art. 312); hence 

m 1 v 1 = m 2 \/2gh, or v t = (m 2 /m l )\ / ^2gh. 

EXAMPLES. 

1 . A body whose weight is W is dropped twice from a height 
h, once striking on a pile of hay and once upon the ground. If 
the times of the impacts are t' and t" , compute the average force 
of the blow in each case. 

2. If a 2-oz. lead bullet strikes a plate with a velocity of 1000 
ft.-per-sec, and is "flattened out" in 1/100 sec, what is the 
average force of the blow ? 

333. Collision or Impact. — Consider two bodies in a collision 
such that the only sudden impulses involved are those of the 
pressures which they exert upon each other. Then the impulse 
of other forces (as gravity) which may act on the bodies during 

* This average, it should be noted, is a time-average, and the average 
force of the chapter on Work and Energy (XIV) is a space-average. 



§ IV.] APPLICATIONS. 345 

collision are negligible and the principles of conservation apply. 
In such a collision there is no change in the momentum or mo- 
ment of momentum of both bodies together. 

Definitions : If the mass-centres of two bodies before collision 
move aloi.g the same straight line, the impact is called direct. 
If the forms of bodies are such that the pressures which they 
exert upon each other are directed along the line joining their 
mass-centres, the impact is called central. 

334. Direct Central Impact. — Let the common path of the 
mass-centres be taken as an x axis, then the momentum of each 
body has no y or z component. Since the impulses due to the 
collision are directed along the x axis, the increment in the 
momentum of each body due to the collision is directed along 
the x axis and the momenta of the bodies after the collision have 
no y or z components, i.e., the mass-centres of the bodies move 
along the x axis after the impact. 

The moment of the impulse of the force exerted upon either 
body about any axis through its mass-centre is zero ; hence the 
moment of momentum of each body about any axis through its 
mass-centre is unchanged in the collision. In particular, if the 
motion of either body before collision is translatory it will be 
so after collision. 

Let A and B be two translatory bodies in collision and let 
w t and m 2 denote the masses of A and B ; 
v 1 and v 2 " " velocities of A and B before impact; 
v x ' and v 2 "A and B after impact. 

The velocities should be regarded as having sign, those in one 
direction being positive and those in the other negative. As 
previously explained, the total momentum of the two bodies is 
not changed by the impact; hence 

m 1 v 1 + m 2 v 2 = m 1 v l ' + m 2 v 2 (1) 

It has been determined experimentally that when two spheres 
collide directly and centrally the velocity of either relative to 
the other is reversed by the impact and diminished, and that the 
diminution depends only on the materials. That is, 

(v x - v,)=-e(v 1 '-v./), (2) 

e being a common fraction and called "coefficient of restitution." 



34& IMPULSE AND MOMENTUM. [Chap. XV. 

We assume that relative velocities of any two bodies before and 
after a direct and central impact are related as in the case of 
spheres. 

Equations (i) and (2) determine the final velocities of two col- 
liding bodies in terms of their masses, initial velocities, and their 
coefficient of restitution. Thus we find from the equations that 

v 1 f = v 1 - (i+e)(v 1 -v 2 )m 2 /(m 1 +m 2 ), ... (3) 
v 2 ' = v 2 -(i+e)(v 2 -v 1 )m 1 /(m 1 +m 2 ). ... (4) 

The value of the impulse equals the change in the momentum 
of either body. This change is 

mpi — m 1 v 1 = — (1 +e)(v 1 — v 2 )m l m 2 /(m l +m 2 ). . (5) 

During a collision each body is first compressed, and after 
full compression has been reached it begins to recover its natu- 
ral form unless the body is perfectly inelastic. The time occu- 
pied by the compression is called the period of compression, and 
it is assumed to be alike for both bodies. The remainder of the 
time of an impact is called the period of restitution. When the 
compression ends and restitution begins the velocities of the 
mass-centres are the same; let v denote this common velocity. 
Then, from the principle of conservation, 

(m 1 v 1 + m 2 v 2 ) = (m 1 + m 2 )v = m{i\' + nt 2 v 2 ; 

w 1 u 1 +w 2 v 2 m x v x ' +m 2 v 2 ' 

hence v = ■ = ; (0) 

m 1 +m 2 m 1 +m 2 

For the case of a body (m^ impinging on a fixed one (m 2 ), 
we substitute for m 2 and v 2 respectively 00 and o ; then 

v 1 / =-ev 1 (7) 

335. Loss of Energy in an Impact. — There is always a loss of 
kinetic energy in an impact (unless e=i), its value being found 
as follows: The kinetic energy before and after impact equals 

\m{v? + \m 2 v 2 2 and \m{v{ 2 + \m 2 v 2 2 . 

Subtracting the latter from the former and substituting for v t * 
and v 2 ' their values from (3) and (4), we have as the loss 

i(l „^2 ) J^ ( Vl -v 2 y. 



§ IV.] APPLICATIONS, 347 

EXAMPLES.* 

. i . Show that if two bodies of equal mass and perfectly elastic 
(e = i) collide, they exchange velocities. 

2. If a ball falls from a height h upon a horizontal plane, 
show that the height of rebound equals e 2 h. 

3. Two bodies of unequal mass with momenta numerically 
equal meet. Show that their momenta after impact are still 
numerically equal. 

4. A body whose mass is 10 lbs., moving with a velocity of 
87 ft.-per-sec, overtakes a body whose mass is 40 lbs., moving 
with a velocity of 6 ft.-per-second. The coefficient of restitu- 
tion being 3/4, compute the final velocity of each and the im- 
pulse of the forces due to impact. Ans. v 2 f = 6.7 ft. /sec. 

5. Suppose that in ex. 4 the bodies meet and collide; then 
solve. 

6. Suppose that in ex. 4 the bodies are inelastic, and solve. 

7. Deduce values of the impulses for the periods of com- 
pression and restitution, and show that the former is independ- 
ent of the coefficient of restitution. 

336. Ballistic Pendulum and Centre of Percussion. — Fig. 
246 represents a ballistic pendulum (for determining the ve- 
locity of a projectile); let M denote its mass, k its | R , 
radius of gyration with respect to its axis of rota- T . " 
tion, m the mass of the shot, v its striking velocity, 
and co the angular velocity produced in the pendu- 
lum by the impact. The velocity of the shot just ! 
after it is imbedded and has come to rest relative 
to the pendulum equals rco. Since the time of the 
impact (during which the angular velocity is gen- vj_ 
erated) is very short, the angular displacement of 
the pendulum during that time is practically zero, FlG - 2 4 6 - 
and the direction of the velocity of the shot just after imbedding 
is practically horizontal. Hence the impulse exerted on the 
shot by the pendulum and that exerted on the pendulum by th< 
shot equal 

m(v — rco). 

* Direct central impact is implied in all the examples. 



c._ 



34§ IMPULSE AND MOMENTUM. [Chap. XV. 

The moment of the impulse of the shot on the pendulum with 
respect to the axis is m(v — rw)r, and the change in the moment 
of momentum of the pendulum (with respect to the axis) dur- 
ing the time of the impulse equals Mk 2 a> (see art. 326). Ac- 
cording to B, art. 328, 

m(v — rco)r = Mk 2 aj, or v = (Mk 2 +mr 2 )a>/mr. 

If h is the height to which the centre of gravity of the pendulum 
rises, (o = (i/k)\/~2gh (see ex. 8, art. 312); hence 

_ Mk 2 + mr 2 ,— 
mrk * ' 

and from this equation v may be computed, the quantities on 
the right side being readily measured in any actual cr.se. 

Centre of Percussion. — Let R f and R" denote the average 
values of the vertical and horizontal components of the hinge 
reaction on the ballistic pendulum during an impact, P the aver- 
age value of the force of the blow, and t the duration of the blow 
which is assumed to be so short that the displacement of the 
pendulum for the time t is practically zero. 

The momentum of the pendulum after the blow is Mr to and 
its direction is horizontal, and the moment of its momentum 
about the axis is Mk 2 co (see art. 326); hence, according to the 
principles of impulse and momentum (see fig. 246), 

R't-Wt = o, 
Pt-R"t = Mrco, 
(Pt)r = Mk 2 co. 

From these, R' = W and R" = Mco(k 2 /r-r)/t. 

Observe that R' is independent of the blow and that R" 
equals zero if the blow is applied at a distance equal to k 2 /r 
below the axis. The point in OC produced, whose distance 
from equals k 2 /r, is called centre 0} percussion; it coincides with 
the centre of oscillation of the pendulum (see art. 267), and the 
methods given in art. 267 may be employed to locate the centre 
of percussion of a ballistic pendulum. 



APPENDIX A. 

VECTORS. 

A i. Scalar and Vector Quantities. — A quantity which has magni- 
tude or magnitude and sign only is called a scalar quantity. An 
amount of money, a volume, etc., are examples of scalar quantities. 

A quantity which has magnitude and direction is called a vector 
quantity. A step, a force, and a velocity are examples of vector 
quantities. 

The methods of ordinary algebra are sufficient for purposes of 
analysis in mechanics when only scalar quantities are concerned, 
but insufficient in general for dealing with vector quantities. For 
example, the algebraic sum of two forces is in general meaningless 
or at any rate without mechanical significance, but their vectorial 
sum (to be explained) has a very important significance. There is 
a branch of mathematics sometimes called Vector Algebra, the 
methods of which are especially adapted for dealing with vector 
quantities. We proceed to a brief explanation of Addition and 
Subtraction by those methods. 

A 2. Vector Defined. — A straight line of definite length and 
direction is called a vector. The word direction here refers not 
only to the inclination (or "clinure") of the line, but also to its 



F -*- 




(a) (b) 

Fig. A i. 

''sense " (" right- or bft-ness " or " up- or down-ness " along the line). 
The sense is usually indicated by an arrow-head placed on the line. 
The lines of fig. A i are vectors. 

Two vectors in order to be equal must be equal in length and 
of the same direction. The first and second vectors of fig. A i (a) 
(counting downwards) are equal because they agree in length and 

349 



35° 



APPENDIX A. 



direction, but the third is not equal to either of the others because 
its direction is different from theirs. 

A 3. Addition of Vectors. — Definition. — The sum of the vectors 
AB and BC is the vector AC (fig. A 1 b). Notice that this defini- 
tion does not conflict (as at first sight it may appear) with the 
proposition of geometry which states that the length of any side 
of a triangle is less than the sum of the lengths of the other two. 

To add more than two vectors we proceed as in the ordinary 
algebra, i.e., we add any two, then to that sum another vector, and 
so on until all have been combined. As in the addition of scalars, 
the sum of several vectors does not depend on the order in which 
they are added; thus the sum of the four vectors a, b, c, and d (fig. 
A 2) is found to be AB by adding them in a certain order and CD 






Fig. A 2. 

by adding in another order, and AB and CD are found to agree in 
length and direction. If they are added in any other order, the 
sum will be found to be equal to A B or CD. 

EXAMPLE. 

Reverse the arrows on vectors c and d (fig. A 2) and add the 
four vectors in at least two ways. 

A 4. Negative of a Vector. — In ordinary algebra, the negative of 
a quantity is one which added to the quantity gives a sum equal 
to zero. So, too, the negative of a vector is one which added to 
the vector gives a sum equal to zero. Now the sum of two vectors 
which are equal in length, parallel, and opposite in sense equals 
zero ; hence either is the negative of the other. 

A 5. Subtraction of Vectors. — In ordinary algebra, a quantity is 
subtracted from another by adding its negative. So, too, to sub- 
tract a vector from another we add the negative of the former to 
the latter. 

EXAMPLES. 

1. Subtract vector a (fig. A 2) from vector b. 

2. Subtract vector b from vector a. 



APPENDIX B. 

RATES.* 

The term rate is in common use, but usually in an inexact sense. 
In the following a precise meaning is given to it which, it will be 
observed, does not conflict with the popular notions in so far as 
they are exact. 

B i. Kinds of Variable Quantities. — Let x and y be two quanti- 
ties which are related to each other, a change in x producing a 
change in y. If all equal changes in x (large or small) produce 
equal changes in y, y is said to vary uniformly with respect to x 
and it is called a uniform -variable. If equal changes in x produce 
unequal changes in y, it is said to vary non-uniformly with respect 
to x, and it is called a non-uniform Variable. 

B 2. Rate of a Uniform Scalar. — If y is a -uniform variable, then 
the locus representing the relation between x and y is a straight 
line, for in such a locus equal changes in x 
produce equal changes in y. 

In this simple case, the popular meaning 
of "the rate of y" is definite, it being the 
change in y per unit change in x. If Ax, 
%2 —%v (see fig. B i,) denotes any change in x 
and Jy, y 2 —y x , the corresponding change in 
y, then the change in y per unit change in x 
is Jy/Jx t or (y 2 —y l )/(x 2 —x l ). Hence if r 
denotes rate, 

r = Jy/Jx (i) 

Evidently r is the same for all values of Jx, i.e., the rate of a uni- 
form variable is constant. 

B 3. Rate of a Non-Uniform Scalar. — If y denotes a non-uniform 
variable, then the locus representing the relation between x and y 
obviously must be curved. 

According to popular notions, the rate of a non-uniform vari- 

* This appendix is intended for students who do not associate the 
idea of rate with dy/dx, but do understand that dy/dx is a "limit." 

3Si 




K ! 
yf Ax \y 2 




v—Xt— * 

« ---War > 


(a). 



Tx 

a! r2 !B 


X 


*— £B T — > 


(6) 



35 2 APPENDIX B. 

able is not constant and has a definite value at each value of the 
variable, but these definite values of the rate are not clearly dis- 
cerned in the popular mind. 

Average Rate Defined. — Let Ay denote the change in y due to a 
change Ax in x (see fig. B 2 a) . Evidently the rate of a uniform vari- 
able, also depending on x, might be 
such that the change in that variable 
due to a change in Ax in % equals Ay. 
The average rate of the non-uniform 
variable for the range x 2 —x x is de- 
fined as the rate of such a uniform 
variable as just mentioned. Now 
the rate of the uniform variable is 
Ay /Ax; hence if r a denotes average 
rate of the non-uniform variable, 

T a = Ay/Ax. ... (2) 

True or Instantaneous Rate De- 
fined. — The value of the average rate 
of a variable is generally different for 
FlG - B 2 * different values of x 2 — x x , but (see 

the figure) the average rate, Ay/ Ax, approaches a definite value as Ax 
approaches zero (x 2 approaches x x , for instance). The rate of the 
variable y at the value y = y t is defined as the limit of the average 
rate of y for a range x 2 —x x as x 2 approaches x x . Now the limit of 
Ay I Ax is dy/dx, i.e., the rate of y at the value y = y x equals the value 
of dy /dx corresponding to y — y x (and x = x l ). Hence if r denotes the 
rate of y at any value of y, 

r = dy/dx . (3) 

Observe that the rate of y at the value y=y t is represented by 
the slope of the tangent to the locus representing the relation be- 
tween x and y at the point x x y v 

Consistency of the Definitions. — The average value of a number 
of quantities is computed by adding them and dividing the sum by 
their number — the quotient being the average sought. It can be 
shown that the above definitions of average and instantaneous rates 
agree with this method of computing averages. Thus let r x and r 2 
denote the values of the rate of y when x = x x and x 2 respectively 
(see fig. B 2 b), and let ordinates to the curve represent values of 
the rate at intermediate values of x. Then since the average ordi- 
nate represents the average rate, the latter is given by 
r a = (area ABB x A t ) /{x 2 -x x ). 



RATES. 



353 



It is shown in works in calculus that the area ABB^A X equals / rdx, 

Jxi 

hence 

x 2 —x l Jx l Xo—x^yi x 2 —x x Ax 

Thus we find that the ordinary method of computing averages leads 
to a result identical with that given by eq. (2). 

The definition of " instantaneous rate " further agrees with popu- 
lar notions, as may be explained thus: Let y and z denote uniform 
and non-uniform variables respectively, both depending on x, and 
suppose that their relations to x are represented by the straight and 
curve loci of fig. B 3. From the figure it is plain that for any 
change Ax between A and B (at A 2 the tan- 
gent to the curve is parallel to the straight 
line) the change, or increment, in z is 
smaller than that in y, and that they be- 
come more nearly equal the nearer B is 
to A. Also for any change Ax between A 
and C the change in z is greater than that 
in y, and they become more nearly equal 
the nearer C is to .4 . Popularly stated : up 
to A , the rate of z is less than the rate of y, 
beyond A the rate of z is greater than that, 
of y, and at .4 they are equal. Now our formulas for rate should 
agree with this popular expression, and it is readily shown that 
they do. Thus the rates of z and y are dz/dx and dy/dx ; and it will 
be seen from the figure that for values of x less than x f the former 
is the lesser, for values of x greater than x' the former is the greater, 
and for x = x f they are equal. 

B 4. Sign of a Rate. — Both Ay I 'Ax and dy/dx maybe negative as 
well as positive; hence in order that equations (1), (2), and (3) may 
be true as to sign, it is necessary to regard a rate as having sign 
and the sign must be the same as that of the expression for the rate 
{Ay J Ax or dy/dx). Now Ay / Ax (for a uniform variable) and dy/dx 
are positive when y increases as x increases, and they are negative 
w T hen y decreases as x increases.* 

Hence the rate of y with respect to x at any particular 
value of y is positive or negative according as y in- 
creases or decreases at that value as x increases. 




* Algebraic increase and decrease are meant. 



354 



APPENDIX B. 



Thus in fig. B i (a) the rate of y is positive, in fig. B i (6) it is neg- 
ative, and in fig. B 2 (a) the rate has different signs at different 
values of x or y, it being positive from m to n, negative from n to o. 
B 5. Unit of Rates. — The expressions for rates in eqs. (1), (2), 
and (3) imply a certain unit, namely, the rate of a uniform variable 
y which changes a unit in amount for a unit change in x. Thus if 
y denotes volume and x distance, the unit rate might be one cubic 
foot per foot, one gallon per inch, etc. ; if y denotes distance and x 
time, the unit rate might be one foot per second, one mile per hour, 
etc., etc. 

B 6. Rate of a Uniform Vector. — Let y denote a vector which is 
related to x and letOa, Ob, Oc, etc. (fig. B 4), represent y at values of 

x equal to x v x 2 , x 3 , etc., the dif- 
ferences between successive values 
of x being equal (x 2 —x i =x 3 —x 2 , 
etc.). Now if ab=bc = cd etc., 
y is a uniform variable, for the 
changes (or increments) in y (vec- 
tors) for equal changes in x are 
equal. 

By rate of a uniform vector y 
is meant its change per unit change 
in x\ hence the rate equals any change in y (a vector) divided by 
the corresponding change in x, as (vector ac) /{x 3 —x v ). If OA and 
OB represent y for any values of x as x and x + Ax respectively, then 
the rate of y is 

(vector AB)/Ax. 



^-zl-S--* 




Fig. B 4. 



Let P be a fixed point in ad in the direction BA from A , and let 
5 denote the variable distance PA. The increment of 5 due to a 
change Jx in x equals A B as shown, and since y is a uniform vari- 
able, 5 evidently is also; hence the rate of 5 is As /Ax. Finally, 
since 5 = length AB, the rate of y is a vector 

whose direction is that of the increment of y and 
whose magnitude equals the rate of s. 

B 7. Rate of a Non- Uniform Vector.— Let Oa, Ob, Oc, etc. (fig. B $), 
represent a vector y at values of x equal to x v x 2 , x z , etc., the 
successive differences in x being equal. The changes in y corre- 
sponding to changes x 2 —x 1 , x 3 —x 2 , etc., are the vectors ab, be, etc. 
Then if, in fig. B 5 (a), ab, be, etc., are unequal and if, in Fig. B 5 (b), 
ab, be, etc., are equal or unequal, y is a non-uniform variable; for 



RATES. 355 

the changes, or increments, in y due to equal changes in x are not 
equal.* 

Average Rate Defined. — The average rate of a vector y for any 
range Ax in x is denned as the rate of a uniform vector whose change, 
for the same range in x, equals that of y. Thus if OA and OB repre- 



ss--* ,-- 






sent y for values x and x + Ax, the change in y for the range Ax 

is the vector AB. Now the rate of a uniform vector whose change 

is also AB for the range Ax is (vector AB)/Ax, and this, by the 

definition, is the average rate of y during the change Ax, i.e., 

the average rate is a vector whose direction is AB and 

whose magnitude equals (length A B) / Ax. 

Actual or Instantaneous Rate Defined. — The value of the average 

rate of y is different in magnitude and in direction for different 

values of x. The average rate approaches a definite direction and 

a definite magnitude as Ax approaches zero. The rate of the 

variable y at the value y = OA is defined as the limit of the average 

rate of y as Ax approaches zero (B approaches ^4). The limiting 

direction of the average rate is the limiting direction of AB, which 

is the direction of the tangent at A. The limiting value of the 

average rate is the limit of (chord AB) / Ax, which is the same as 

the limit of (arc AB) / Ax. Now let P be any fixed point on the 

curve A B in the direction B A from A, and 5 the distance from P 

to A ; then the change in s, As, due to a change A x in x is the arc 

AB. Hence 

,. arc AB -. ..As ds 

lim = limit — =-r-. 

Ax Ax dx 

Finally, the rate of y at the value y = OA is a vector 

whose direction is that of the tangent at A , and 
whose magnitude = ds /dx. 

* Equal vectors are equal in length and the same in direction. 



35 6 APPENDIX B. 

It is implied that the fixed point from which 5 is measured is 
so taken that As is positive. The arrow on the tangent, giving the 
sense of the rate, points in the positive 5 direction. 

B8. Descriptive Terms. — A variable y may depend on several 
quantities. If so, it has as many rates, and we say for brevity the 
" xvoXe of y " to distinguish the rate of y with respect to x from its 
other rates. The rate dy/dx is often described by naming the kind 
of a quantity represented by x\ thus if x denotes distance, dy/dx is 
called a "space rate" of y; if x denotes time, dy/dx is called a 
" time rate "of y. The time rate of y is sometimes indicated 
thus, y. 

If y and x denote certain quantities, the rate is given a single 
name; thus if y and x denote distance and time respectively, dy/dx 
is called velocity; if y denotes mass and # volume; dy/dx is called 
density. 



APPENDIX C. 

DIMENSIONS OF UNITS. 

C i. Magnitude of a Quantity. — The magnitude of a quantity is 
expressed by stating how many times larger it is than a standard 
quantity of the same kind and naming the standard. Thus, we 
say that a certain distance is 10 miles, meaning that the distance is 
10 times as great as the standard distance, the mile. 

The number expressing the relation between the magnitude of 
the quantity and the standard (the number 10 in the illustration) 
is called the numeric (or numerical value) of the quantity, and the 
standard is called the unit. 

C 2. Fundamental and Derived Units. — A unit for measuring any 
kind of quantity may be selected arbitrarily, but it must of course 
be a quantity of the same kind as the quantity to be measured 
(e.g., a unit for measuring lengths must be a length). Thus, as 
unit of velocity we might select the velocity of light, as unit of area 
the area of one face of a silver dollar, etc. Many units in use are 
arbitrarily chosen, i.e., without reference to another unit (e.g., the 
bushel and the degree), but it is convenient practically to define 
them with reference to each other. All mechanical and nearly all 
physical quantities can be defined in terms of three arbitrarily 
selected units, i.e., ones not dependent on any other units. These 
are called fundamental units, and the others, defined with reference 
to them, derived units. It is customary in works on theoretical 
mechanics and physics to choose as fundamental the units of 

length, mass, and time, 

but it is sometimes more convenient to take as fundamental the 
units of 

length, force, and time. 

In the following article we give a discussion of derived units with 
reference to each of these sets of fundamentals, and on pages 360 and 
361 there appear summaries in which the absolute units are re- 
ferred to the first set of fundamentals and the gravitational units 
to the second set. But either set might serve as fundamentals for 
all absolute and gravitational units. 

357 



35 8 APPENDIX G 

C 3. "Dimensions" of Units. — A statement of the way in which 
a derived unit depends on the fundamental ones is called a state- 
ment of its dimensions. 

Obviously an area depends only on the unit of length, definitely 
as the square of the unit of length. Thus 

(one sq. yd.) /(one &q. ft.) = (one yd. or three ft.) 2 /(one ft.) 2 = 9. 

This relation is expressed in the form of a " dimensional " equation, 
thus 

(unit area) = (unit length) 2 , 

and briefly a unit area is said to be "two dimensions in length." 
Similarly a unit volume is said to be three dimensions in length. 

Velocity. — According to the definition of velocity (art. 167), a 
unit velocity is directly proportional to the unit length and in- 
versely to the unit time; hence if V, L, end T denote units of veloc- 
ity, length, and time respectively, the dimensional equation is 

V=L/T=LT- 1 , 
and a unit velocity is one dimension in length and minus one in time. 

Acceleration. — According to the definition of acceleration (art. 
173), a unit acceleration is proportional directly to the unit veloc- 
ity and inversely to the unit of time ; hence if A denotes unit accel- 
eration, the dimensional equation is 

A=V/T=L/T 2 =LT- 2 , 

and a unit acceleration is one dimensiomin length and minus two 
in time. 

Angular Velocity. — According to the definition of angular veloc- 
ity (art. 210), a unit angular velocity is proportional directly to 
the unit of angle and inversely to the unit of time ; hence if co and 
denote units of angular velocity and angle respectively, the dimen- 
sional equation is 

(u = 0/T, or co^T- 1 , 

since units of angle, (degree, radian, etc.) are independent of the 
fundamental units. A unit angular velocity is therefore minus one 
dimension in time. 

Angular Acceleration. — According to the definition of angular 
velocity (art. 213), a unit angular acceleration is proportional 
directly to the unit angular velocity and inversely to the unit time ; 
hence if a denotes unit angular acceleration, the dimensional equa- 
ion is 

and a unit angular acceleration is minus two dimensions in time. 



DIMENSIONS OF UNITS. 359 

Force. — In accordance with the equation of motion of a particle 
(art. 236), R=ma, or 

" force = mass X acceleration ", 

i.e., the unitforce is directly proportional to the units of mass and 
acceleration; hence if F and M denote units of force and mass 
respectively, the dimensional equation is 

F=MA=LMT- 2 , 

and a unit force is one dimension in length, one in mass, and minus 
two in time. 

Mass. — If we regard length, force, and time as fundamental 
units, then the last equation written as follows is the dimensional 
equation for a unit mass : 

M=FT 2 /L=L- 1 FT 2 , 

and a unit mass is minus one dimension in length, one in force, and 
two in time. 

Work. — According to the definition of work (art. 286), the unit 
of work is directly proportional to the units of force and length; 
hence if W denotes unit work, the dimensional equation is 

W=LF=L 2 MT- 2 , 

and a unit work is one dimension in length, one in force, or two in 
length, one in mass, and minus two in time. 

Power. — According to the definition of power (art. 311), a unit 
of power is proportional directly to the unit of work and inversely 
to the unit of time ; hence if P denotes unit of power, the dimen- 
sional equation is 

p = w/T =LFT~ 1 =L 2 MT- 3 , 

and a unit power is one dimension in length and force and minus one 
in time, or two in length, one in mass, and minus three in time.* 

C 4. Applications of the Theory of Dimensions. — A knowledge of 
the theory of dimensions is probably of most value to the beginner 
as a help to a clear understanding of the different mechanical quan- 
tities and the relations between them. The theory is useful prac- 
tically in other ways, two of which we mention. 

(1) As a test of the accuracy of equations between mechanical quan- 
tities. — Such an equation if rationally and correctly deduced must 
be homogeneous, i.e., the terms in it must be the same in kind. To 

* Determination of the dimensions of the other units mentioned in the 
following tables is left to the student. 



360 



APPENDIX C. 



ABSOLUTE SYSTEMS ("SCIENTIFIC"). 



Names of Quantities. 



Length 

Mass 

Time 

Velocity 

Acceleration 

Angular Velocity 

Angular Acceleration. . . . 

Force 

Weight 

Moment of Mass 

Moment of Inertia (Body) 

Moment of Force 

Work 

Energy 

Power 

Impulse 

Momentum 

Density 

Specific Weight 

Moment of Area 

Moment of Inertia (Area) . 

Stress 

Stress Intensity. ........ 



Dimen- 
sional 
Formulas. 



L 

M 

T 

LT- 1 

LT~ 2 

T- 1 

LMT- 2 

LMT-' 1 

LM 

L 2 M 

L 2 MT~ 2 

L 2 MT~ 2 

L 2 MT~ 2 

L 2 MT~ 3 

LMT- 1 

LMT- 1 

L~ 3 M 

L- 2 MT- X 

L 3 

-L 4 

LMT- 2 

L-'MT- 2 



Names of Units. 



C.G.S. 



centimeter (cm) 

gram (gr) 

second (sec) 

cm /sec ("kine") 

cm /sec 2 ("spoud") 

rad/sec 

rad/sec 2 

dyne 

dyne 

gr-cm 

gr-cm 

cm-dyne 

cm-dyne ("erg") 

cm-dyne ("erg") 

erg /sec 

dyne-sec ("bole") 

dyne-sec ("bole") 

gr/cm 3 

dyne /cm 3 

cm 3 

cm 4 

dyne 

dyne /cm 7 



F.P.S. 



foot (ft) 

pound (lb) 

second (sec) 

ft /sec 

ft/sec ! 

rad/sec 

rad/sec* 

poundal (pdl) 

pdl 

lb-ft 

lb-ft 

ft-pdl 

ft-pdl 

ft-pdl 

ft-pdl /sec 

pdl-sec 

pdl-sec 

lb/ft 3 

pdl/ft 8 

ft 3 

ft* 

pdl 

pdl/ft 8 



ascertain whether terms are the same in kind we write the dimen- 
sional form of the equation, reduce the terms to their simplest 
forms and compare; if they are alike, the terms are the same in 
kind. To illustrate, consider eq. (1), art. 250, 



g+$+rt"-«*ltfn-. 



(1) 



in which y and A denote lengths, / and 1 /p time, and q and at angu- 
lar velocity. Then d 2 y/dt 2 is an acceleration and dy/dt a velocity, 
and the dimensional equation is 

LT- 2 + (T- 1 )(LT- 1 )+(T- 1 ) 2 (L)=(T- 1 ) 2 L. 

An abstract number, as the sine of an angle, is independent of all 
units and hence does not affect a dimensional equation. Reducing 
the terms in the last equation we get 

LT- 2 +LT- 2 +LT- 2 =LT- 2 ; 

i.e., the terms are alike and the original equation is homogeneous. 
Showing that the equation is homogeneous does not prove that 



DIMENSIONS OF UNITS 



361 



GRAVITATION SYSTEMS ("ENGINEERS'"). 



Names of Quantities. 



Dimen- 
sional 
Formulas. 



Names of Units. 



F.P. (force) S. 



M.K. (force) S. 



Length 

Force 

Time 

Velocity 

Acceleration 

Angular Velocity. . . . 
Angular Acceleration 

Mass 

Weight 

Moment of Mass. . . . 
Moment of Inertia. . . 
Moment of a Force. . 

Work 

Energy 

Power 

Impulse 

Momentum 

Density 

Specific Weight 

Moment of Area. . . . 
Moment of Inertia. . . 

Stress 

Stress Intensity 



L 

F 

T 

LT- 1 

LT- 2 
T -x 

T~ 2 

L~ X FT 2 

F 

FT 2 

LFT 2 

LF 

LF 

LF 

LFT- 2 

FT 

FT 

L 2 FT 2 

L-*F 

L 3 

L 4 

F 

L~ 2 F 



foot (ft) 
pound (lb) 
second (sec) 
ft /sec 
ft/sec 2 
rad/sec 
rad/sec 2 
'geepound" (g 
lb 
glb-ft 
glb-ft 2 
ft-lb 
ft-lb 
ft-lb 
ft-lb /sec 
lb-sec 
lb- sec 
glb/ft 3 
lb/ft 3 
ft 3 
ft 4 
lb 
lb/ft 2 



lb) 



meter (m) 

kilogram (kg) 

second (sec) 

m/sec 

m/sec 2 

rad/sec 

rad/sec 2 

'geekilogram" (gkg) 

kg 

gkg-m 

gkg-m 2 

m-kg 

m-kg 

m-kg 

m-kg/sec 

kg-sec 

kg-sec 

gkg/m 3 

kg/m 3 

m 3 

m 4 

kg 

kg/m 2 



it is correct, but that it may be correct ; showing that an equation is 
non-homogeneous shows it to be incorrect. Since abstract num- 
bers do not appear in the dimensional form of an equation, the test 
for homogeneity does not discover errors in numerical coefficients 
and terms, nor of course errors in signs. 

As another illustration consider eq. (2), art. 250, 



A sin (o^+e), 



(2) 



It was deduced from eq. (1) by two integrations. If we find that 
(2) is non-homogeneous, we may be sure that a mistake was made in 
the deduction from (1) to (2). The dimensional form of (2) is 



(TV 



(T-*)(T->) 



L=L, 



i.e., the equation is homogeneous and it may be correct. 

Not only must rational mechanical equations be homogeneous, 
but every factor or expression in it which is the sum of several terms 
must also be homogeneous. For example, in eq. (2) the expression 



362 APPENDIX C 

{at + e) must be homogeneous, and since e denotes an angle, it is 
readily seen to be so. 

(2) To express a magnitude in different units. — Obviously the 
numerical value of a given quantity changes inversely as the mag- 
nitude of the unit used ; thus a certain distance may be expressed as 

10 mi., 17,600 yds., and 52,800 ft., 

and plainly the numerics are respectively as 1, 1760, and 5280, while 
the corresponding units are as 5280, 1760, and 1. 

Let q x be the known numerical value of a quantity when ex- 
pressed in the unit Q v and q 2 the numeric (to be found) of the 
same quantity expressed in the unit Q 2 ; then 

( h=9± or a =aQ* 

The ratio Q l /Q 2 can be easily computed by substituting for Q t and 
Q 2 their equivalents in terms of fundamental units; thus if a, b, and 
c are the dimensions of Q x (and Q 2 ), 

Q X = K{L%M\T\) and Q 2 = & 2 (L^r 2 ), 

where L 1 ,M 1 , and 7\ are the particular fundamentals for Q v L 2 , M 2 , 
and T 2 those for Q 2 , and k x and k 2 numerical coefficients (very often 
unity). Finally, 

As an example, let us determine how many watts in 10 horse- 
power. Since Q x (horse-power) =550 ft.-lb.-sec. -1 and Q 2 (watt) = 
10' ergs per sec. = io 7 cm. -dyne-sec. -1 , 



sec - 



550 ft. lb. 

io 7 cm. dyne sec -1 

= io^£ (30.48X4.45 Xio 5 )(i) = 76 4 o. 



APPENDIX D. 

SECOND MOMENTS OF AREAS (MOMENT OF INERTIA, ETC.).* 

In the subject of strength of materials especially, the student of 
engineering meets with quantities expressed by integrals of the kind 

and form / dA • x 2 and / dA • xy, A denoting area and x and y dis- 
tance. Such quantities have been called ''moments of area of the 
second order," or briefly " second moments of area," the terms being 
in line with " first moments of area," which term is applied to quan- 
tities expressed by integrals like / dA>x (see art. 83). We distin- 
guish between second moments of area employing special names 
for the kinds. 

§ I. Moment of Inertia. 
D 1. Moment of Inertia Denned. — The moment of inertia of a 

plane area with respect to any axis is the sum of the products 
obtained by multiplying each elementary part of the area by the 
square of its distance from the axis. 

The axis of reference will often be called ' ' inertia-axis ' ' to dis- 
tinguish it from other axes, coordinate, geometrical, etc. We con- 
sider only moments of inertia with respect to axes in or normal to 
the plane area; the latter are called polar moments of inertia, and 
the corresponding axes polar axes. 

Expression for Moment of Inertia. — Let dA v dA 2 , dA 3 , etc., de- 
note elementary parts of an area and p v , p 2 , p 3 , etc., respectively 
their distances from some axis; then according to the definition, 
the moment of inertia of the area with respect to that axis is 

(dA l ) Pl * + (dA 2 )p 2 > + ... 

Or, if 7 t denotes the moment of inertia, dA any elementary por- 

* Writers on Strength of Materials usually refer to works on 
Mechanics for a treatment of these second moments, and for that reason 
this appendix is herein included. 

f A subscript affixed to the symbol refers to the inertia-axis; thus I \ 
stands for moment of inertia with respect to the x axis. 

363 



364 APPENDIX D. 

tion of the area all points of which are equally distant from the 
axis, and p that distance, 



JdA-p 2 . 



[The term moment of inertia of an area is unfortunate because 
beginners are prone to seek reasons for its appropriateness which 
do not exist. They should recognize at the outset that an area 
has no inertia and hence, in the ordinary sense of the words, no 
moment of inertia. The reason why this second moment of an 
area was so called lies in the fact that the moment is closely analo- 
gous to another quantity (a second moment of mass, see art. 254) 
which had previously been called moment of inertia of a body.] 

D 2. Units of Moment of Inertia. — Each term in the preceding series 
is the product of four lengths ; hence a moment of inertia of an area 
is four " dimensions " in length. The numerical value of a moment 
of inertia of an area is usually computed with the inch as unit 
length, and the corresponding unit moment of inertia is called a 
"biquadratic inch," abbreviated thus: in. 4 

D 3. Radius of Gyration. — Since a unit moment of inertia of 
an area is four " dimensions " in length, it can be expressed as the 
product of an area and a length squared. It is sometimes con- 
venient to so express it. 

Definition. — The radius of gyration of an area with respect to 
an axis is such a length whose square multiplied by the area equals 
the moment of inertia with respect to that axis. That is, if k and 
I denote the radius of gyration and moment of inertia of an area A 
with respect to the same axis, 

k 2 A=I, or k = vT/A. 

The square of the radius of gyration of an area with respect to 
an axis is the mean of the squares of the distances of all the equal 
elementary parts of the area from that axis. For let p v p 2 , etc., 
be the distances from the elements (dA) to the axis, and let n de- 
note their number (infinite) ; then the mean of the squares is 

W+P 2 2 +P 3 2 + • • .)/n = (p l 2 dA+p 2 2 dA+ . ..)/ndA=I/A. 

But I /A is the square of the radius of gyration, hence, etc. 

[Like moment of inertia, the term radius of gyration when ap- 
plied to areas is strictly inappropriate. Its use in this connection 
is justified by analogy, the quantity k being closely analogous to 
another quantity which had been previously called radius of gyra- 
tion (art. 255).] 



SECOND MOMENTS OF AREAS. 



365 



4" 



\y 



Fig. D 1. 



EXAMPLES. 

i. Show that the moment of inertia and radius of gyration of a 
rectangle with respect to a central axis parallel to the base are respect- 
ively 

j\ba 3 and 0V1/12, 
b and a denoting the base and altitude respectively. 

Solution: We will take a horizontal strip as 
elementary area (see fig. Di); then dA =bdy, and 

/+0/B 
bdy.y 2 = T \ba 3 . 
-a/2 

Also, A=ba, therefore r 2 -fe=ba 3 /ba, orr=aVi/i2. 

2. Show that the moment of inertia and radius 
of gyration of a triangle with respect to a central 
axis parallel to the base are respectively 

•sVfra 3 and a's/i/iS, 
b and a denoting the base and the altitude respectively. 

Solution: We will take a horizontal strip as elementary area 
(fig. D 2), calling the length of any strip u, then 
dA =udy; also, 

u = l_a_ 
b 1 

Hence 

/aa/3 
(2/3 -y/a)bdyy 3 =?\ba 3 . 
-a/ 3 

Also, since A =%ba, r = ?\ba 3 /%ba, or r = a / s/i/i8. 

3. Show that the moment of inertia and radius of gyration of a 
circle with respect to a diameter are respectively 

\nr k and \r, 
r denoting radius of the circle. 

Solution: We will take a horizontal strip as elementary area 
(see fig. D 3 a) ; then if u denotes the length of any strip, 

dA =udy = 2 Vr 2 — y 2 dy. 
Hence I x = 2 f Vr 2 — y 2 dy -y 2 = %nr* ; 

and since A = nr 2 , 

k x = ^7rr 4 /7rr 2 , or k 2 = \r. 

4. Show that the moment of inertia and radius of gyration of a 
circle with respect to a central polar axis are respectively 

\nr and rVi/2. 




-2-; therefore dA = (2/$—y/a)bdy. 



366 



APPENDIX D. 



Solution: We choose as elementary area one as represented in 
fig. D 3 (6) ; then dA = pdd, and 



Iz= f 2n f\pdd-dp)p 2 = \nr\ etc. 

t/O t/o 



In the preceding examples, the inertia-axes are central, but mo- 
ments of inertia of such areas can be readily computed by integra- 




Fig. D 3. 

tion for inertia-axes which are simply situated with reference to a 
line of the figure. Such cases are rectangle and triangle with inertia- 
axis parallel to a side, circle with inertia-axis parallel to a diam- 
eter, etc. In the next article it is shown how to determine moments 
of inertia without integration in these and similar cases, but, to test 
his understanding of the integration method, the student should 
determine the moment of inertia of 

(a) a rectangle with respect to a side ; 

(b) " triangle " " " " " ; 

(c) " circle " " tangent. 

D 4. Relations between Moments of Inertia and between Radii of Gyra- 
tion with Respect to (a) Two Parallel Axes, (b) Three Rectangular Axes. — 
Proposition I. — The moment of inertia (7) of an area with respect 
to any axis equals its moment of inertia (I) with respect to a parallel 
centroidal axis plus the product of its area (^4) and the square of 
the distance (d) between the axes, or, symbolically, 

1=1 +Ad 2 (1) 

Proof: (a) The inertia-axis is in the plane area. Let the area 
be that represented in fig. D 4 (a), U being the inertia axis and C 
the centroid. Then 

I = fdA-v 2 = fdA(y+d) 2 = fdAy 2 +2dfdA-y+d 2 fdA. 
But fdA -y 2 =I, JdA • y = Ay = o, and J dA -A ; hence, etc. 



SECOND MOMENTS OF AREAS. 



3 6 7 



(6) The inertia-axis is normal to the area. Let the area be that 
represented in fig. D 4 (b) , O the point where the inertia-axis pierces 
the plane of the area, and C the centroid. Then 

/ = fdA(y 2 +x+d) = f dA(y 2 +x 2 ) +2d JdA-x + d 2 fdA. 
Now 
JdA{y 2 +x 2 )=T, [dA-x = Ax = o t and fdA=A; hence, etc. 

Corollary: Dividing both sides of eq. (1) by A, we have 

I/A=7/A+d 2 , or k 2 =k 2 +d 2 , .... (2) 
k denoting the radius of gyration of the area with respect to any axis, 
and k that with respect to a parallel centroidal axis. 





Fig. D4. 

Equations (1) and (2) show that the moment of inertia and 
radius of gyration of an area with respect to a centroidal axis are 
respectively less than those for any other parallel axis. Also, that 
with respect to any axis the radius of gyration (k) is always greater 
than the ordinate (d) from that axis to the centroid, but if k is small 
compared to d> 

k—d and I = Ad 2 , approximately. 

Proposition II. — A polar moment of inertia of an area (I 2 ) equals 
the sum of moments of inertia (I x and I y ) of the area with respect to 
any rectangular axes in the area which intersect the polar axis, or 

l z =lx+ly. (3) 

Proof: In accordance with the notation above, the inertia-axes 
in the plane of the area must be called x and y coordinate axes and 
the polar one the z axis. Then the distance of any point of the 
area from the z axis is V x 2 +y 2 , and therefore 

I* = fdA(x 2 +y 2 ) =fdA-x 2 +jdA>y\ 

But J dA'X 2 =I y and J dA-y i = I x ; hence, etc. 



3 68 APPENDIX D. 

Corollary. — Dividing both sides of eq. (3) by A, we get 

I e /A=Ix/A+I y /A, or k|=ki+kj, . . . (4) 

k s denoting the radius of gyration with respect to the polar axis, 
k x and k y those with respect to any rectangular axes in the plane of 
the area cutting the polar one. 

Equations (3) and (4) show that the sums I x +I y and k%+kl 
are the same for all directions of the axis x (and y). Hence if we 
imagine the x and y axes (90 apart) to turn about the polar axis, 
when I x and k x reach a maximum or minimum value, I y and k v are 
a minimum or maximum. 

EXAMPLES. 

1 . Show that the moment of inertia and radius of gyration of a 
rectangle with respect to its base are respectively 

%ba 3 and aVi/3, 

b and a denoting the base and altitude (see ex. 1, art. D 3). 

2. Show that the moment of inertia and radius of gyration of a 
triangle with respect to its base are respectively 

^3 ba 3 and oVi/6, 

b and a denoting base and altitude respectively (see ex. 2, art. D 3). 

3. Show that the moment of inertia and radius of gyration of a 
square with respect to a central polar axis are respectively 

|6 4 and b\/T/6, 

b denoting the length of its sides. 

4. From the result of ex. 3, art. D 3, deduce without integration 
the expression for the moment of inertia of a circle with respect to 
a central polar axis. 

5 . Show that the moments of inertia of a square with respect to 
a diagonal and a central axis parallel to any side are equal. 

6. Compute the moments of inertia and radii of gyration of a 
rectangle 2 X 12 in. with respect to axes 7 inches from the centre and 
parallel to the sides. Compare the radii of gyration with the dis- 
tance from the axes to the centre. Ans. Greater 1 = 732 in. 4 

D 5. Composite Areas. — We refer now to areas which can be 
divided into simple component parts ; e.g., a trapezoid divisible into 
two triangles, a circular annulus, consisting of a circle minus a 
smaller one, etc. The moment of inertia of such an area with 
respect to any axis can be computed by adding algebraically the 



SECOND MOMENTS OF AREAS. 



3 6 9 




•T 

~"if~- 





«— ~6„- 
(6) 

5- 



moments of inertia of its parts with respect to the same axis, the 
moments of inertia of the " negative component parts " being given 
the minus sign. 

EXAMPLES. 

i. Show that the moment of inertia and radius of gyration of a 
circular annulus with respect to a 

diameter (fig. D 5 a) are respect- <—&,--> 

ively 

\(r 2 * -V) and Wr? +V- 
Solution : The moment of iner- 
tia of the larger circle is \-rtr^ \j- -^y 

(see ex. 3, art. D 3) and that of 

the smaller is \nrf. Hence the 

moment of inertia of the annulus Fig. D 

is \x{r 2 k — r x *), etc. 

2. Show that the radius of gyration of a circular annulus with 
respect to a central polar axis is V (r 2 2 -f-r t 2 )/2. 

3 . Show that the moment of inertia and radii of gyration of the 
hollow rectangle with respect to a central axis parallel to the base (fig. 
D 5 b) are respectively 

Ts(^2 a 2 3 ~ ^i a i 3 ) an d [(b 2 a 2 z —b 1 a l 3 )/i2(b 2 a. J , —bjO^f, 

4. Compute the moment of inertia of the "angle" section of fig. 
D 6 with respect to the x and y axes respectively. 

Solution: Consider that the section consists of the rectangle 
A BCD minus the rectangle A'B'C'D. The moments of inertia of 
these with respect to the line CD are (see ex. 1, art. D 4) 

I 3.5 X7 3 = 400.16 and \ 2.5 X6 3 = 180 in. 4 

Hence the moment of inertia of the section with respect to CD is 
400.16—180 = 220.16. The area of the section being 9.5 in. 2 , the 
moment of inertia sought is (see eq. (1), art. D 4), 

220.16 -9.5(7 -2.7i) 2 = 45.32 in.* 

5. Show that for the Z section of fig. D 6 

I x = ^[ba 3 -(b -t)(a -2ty]. 

6. Deduce an expression for the moment of inertia of the T 
section of fig. D 6 with respect to a central axis parallel to the 
base. 



37o 



APPENDIX D. 



7. Compute the moment of inertia of the area represented in 
fig. D 7 (a section of a "built-up" steel beam) consisting of a 
"web plate," two "side plates," and four Z bars) with respect 



£- 9 -iP 



c 1 

i 
i 


c' 


| 


-4 




1 
1 

i 
1 


1 


B' 


A'! 


'A 




A 



X 



<€ 


-&— 


--# 








! 


t 
1 




XT 

1 




1 

1 




1 


*-*-» 


X* 




p 




! 




1 




^ 









(a) 



(6) 
Fig. D 6. 



1 

1 
1 
1 
1 
1 

1 


t i ■ 


iv 


1 ^ 


y 



< 6- 

(C) 



to a horizontal central axis, the moment of inertia of one Z section 
with respect to its horizontal central axis being given as 50.22 in. 4 

(Such numbers and similar data are ob- 

<. is4- tainable from " handbooks " published 

by steel-manufacturing companies.) 

Solution: We compute the moment 
of inertia of the parts separately. Side- 
plate section: moment of inertia with 
respect to its horizontal central axis = 
tV 18 X i 3 = 1.5 in. 4 ; area = 18 in. 2 ; 
distance from its centroid to the inertia- 
axis specified = 7 . 5 in. Hence for two 
side plates 

L-2(i.5+i8X7.5 2 ) = 2028 in. 4 

Z section: moment of inertia with 




Fig. D 7. 



respect to its horizontal central axis = 50.22 in. 4 ; area = 1 o . 1 7 in. 2 ; 
distance between its centroid and the inertia-axis = 3.75 in. 4 ; hence 
for the four Z bars 

7 x = 4 (5o.22 +10.17 X3.75 2 ) =772.84 in. 4 
Web-plate section : moment of inertia with respect to the x axis is 

T2 10 X i 3 = 0.83 in. 4 
For the entire section: 

7x = 2028 +772.84 +0.83 = 2801.67 in. 4 



SECOND MOMENTS OF AREAS. 37 1 

8. The moment of inertia of one Z section (fig. D 7) with 
respect to its vertical central axis being 19.21 in. 4 , compute 
the moment of inertia of the composite area with respect to the 
y axis. 

9. The radii of gyration of the "angle" of fig. D6 with re- 
spect to its horizontal and vertical central axes are 2.19 and 
0.89 in. respectively. Compute the radii of gyration of a pair 
of such sections with respect to their horizontal and vertical central 
axes, they being placed so that AB is an axis of symmetry for the 
pair. 

§ II. Product of Inertia. 

D 6. Product of Inertia Denned. — The product of inertia of a 
plane area with respect to a pair of coordinate axes in the plane is 
the sum of all the products obtained by multiplying each ele- 
mentary area by its coordinates. 

Expression for Product of Inertia. — Let dA v dA 2 , dA 3 , etc., de- 
note elementary parts of an area and (x v yj, (x 2 , y 2 ), etc., their 
coordinates respectively; then, according to the definition, the prod- 
uct of inertia of the area with respect to the coordinate axes is 

(dA l )x 1 y 1 +(dA 2 )x 2 y 2 +etc, 

or, if / * denotes the product of inertia, dA any element of the area 
(dA must be of the second order, as dx dy) , and x and y its coordi- 
nates, then 



/ dA-xy, 



the limits of integration being assigned so that the products dAxy 
for all elements are included in the integration. 

D 7. Units of Product of Inertia. — It is plain from the definition 
and expression of the preceding article that a unit product of inertia 
is four " dimensions " in length. Like moments of inertia we wilJ 
express products of inertia in biquadratic inches. 

EXAMPLES. 

1. Deduce an expression for the product of inertia of the rect- 
angle (fig. D 8) with respect to the coordinate axes. 

Solution : / = J dA • xy = j J°(dx dy)xy = ib 2 a 2 = \A \ 

* When it is necessary to specify the axes with respect to which the 
product of inertia is taken, they are indicated by subscripts thus: J xy . 



372 



APPENDIX D. 




2. Deduce expressions for the products of inertia of the rect- 
angle with respect to coordinate axes through 
2 , . . . 6 and 7 . 

D 8. Zero Products of Inertia. — Unlike a 
moment of inertia, a product of inertia may- 
be negative or zero. The pair of axes pass- 
ing through a given point with respect to 
which the product of inertia of an area equals 
zero is often of practical importance ; in art. 
D 14 a method is given for finding such axes. 
The following proposition enables one to find 
Fig. D 8. those axes more readily in certain cases. 

Proposition. — If an area has an axis of symmetry, its product 
of inertia with respect to that axis and any other perpendicular to 
it is zero. 

Proof: Let the elementary parts of the area be grouped into 
pairs, the elements of each pair being symmetrically situated with 
reference to the axis of symmetry. Obviously the product of iner- 
tia of each pair is zero, and hence the product of inertia of all the 
pairs (the entire area) is also zero. 

EXAMPLES. 

1 . With respect to which axes are the products of inertia of the 
following zero: (1) rectangle; (2) isosceles triangle; (3) circle: 
(4) <" channel," " tee," and " angle of equal legs " (fig. 49)? 

D 9. Relation between Products of Inertia for Parallel Pairs of Axes. 
Proposition. — The product of inertia (J) of an area with respect to 
any pair of coordinate axes equals its product of inertia (/ ) with 
respect to a parallel pair of central axes plus the 
area (A) times the product of the coordinates of 
the centroid {x, y) referred to the first set of 
axes, or 

J=J+Axy (1) 

Proof: Let OX and OY (fig. D 9) be the 
axes and C the centroid. Then 

J=ldA'xy t and J=idA-uv. 

Since x = u +x and y = v +y, Fig. D 9. 

J = JdA {u +x) (v +y) = JdA -uv+y f udA +x JvdA +xyj dA. 

But / udA =uA =0, / vdA =vA=o, and j dA=A; hence, etc. 




SECOND MOMENTS OF AREAS. 



373 



|Y 



D io. Composite Areas. — The products of inertia of such an area 
with respect to any pair of coordinate axes equals the algebraic sum 
of the products of inertia of its component parts with respect to 
the same axes. 

examples. 

i . Compute the product of inertia of the angle section of fig. D 6 
with respect to axes OX and OY. 

Solution: Divide the figure into two rectangles as shown in fig. 
D io; the areas of the parts are 3.5 and 6 in., and 
their centroids are at C x and C 2 respectively, C 
being the centroid of the entire figure. 

For the first rectangle the product of inertia 
with respect to horizontal and vertical axes 
through C\ equals zero (see prop., art. D 8), and 
hence, according to the preceding article, 

Jxy = o +3-5(o.79)( -2.21) = -6. 11 in. 4 
For the second rectangle the product of in- 
ertia with respect to horizontal and vertical axes 
through C 2 equals zero, and hence 

Jxy = o +6( -o. 46)0.29) = -3.56 in. 4 
For the entire figure, therefore, 

Jxy = { -6. 11) +( -3.56) - -9.67 in. 4 
2. Show that the product of inertia of the Z section of fi: 



Co| 



0.46* 



Q -1.29" 

■4 



hw * 



Fig. D io. 



D6 
a, b, and t being 6, 



with respect to the x and y axes is — 11.48 in. 
3^, and I in. respectively. 

§ III. Relation between Moments of Inertia with Respect 
to Axes Inclined to Each Other. 

D 11. General Equations. — Let OX and OY (fig. D 11) be any 
two rectangular axes and OU and OV another pair, XOU being any 

angle a. From the figure it is plain that 

u =y sin a +x cos a, 
and v=y cos a—% sin a. 

If these values for u and v be substituted 

in 

Fig. D 11. Iu-fdA-v 2 and J uv = fdA-uv, 

it will be found that 




374 APPENDIX D. 

Iu — Ix cos 2 a -i-Iy sin 2 a — J X y sin 2a. . . . . (1) 
and Juv = Wx— Iy) sin 2a+/x2/COS ia (2) 

With these equations it is possible to find the moment of inertia 
with respect to any axis through any point and the product of* iner- 
tia with respect to any pair of axes through that point, if the mo- 
ments and the product of inertia of the area with respect to two 
rectangular axes through the point are known. 

EXAMPLES. 

1. Take u and v axes through in fig. D 6 (a) inclined at 45 
and 1 3 5 to the x axis, and compute I u a,ndJ U v, it being given that 
I x , Iy, and J X y equal 45.37, 7.53, and —9.67 in. 4 respectively. 

Solution: Substituting in eqs. (1) and (2), we find that 

^ = 45-37 cos 2 45° + 7.53 sin 2 45 +9.67 sin 90 = 36.12 in. 4 
and /«w = i(45-37 — 7-53) sin 90 —9.67 cos 90° = 18.92 in. 4 

2. Consider a rectangle whose base and altitude equal 4 and 6 
in. respectively, and compute its moment of inertia with respect to 
a line perpendicular to either diagonal at its end; also its product 
of inertia with respect to that line and the diagonal. 

D 12. Geometrical Constructions. — Equations (1) and (2) of the pre- 
ceding article can be solved graphically. There are two graphical 
methods; in one a certain ellipse, the " inertia-ellipse," is employed, 
and in the other a certain circle, which will herein be called " inertia- 
circle." The former possesses more elegance and is probably more 
powerful, but the latter is as much simpler to apply as is the draw- 
ing of a circle compared to that of an ellipse. Only a brief discus- 
sion of the inertia-circle method can be given herein.* 

D 13. Inertia-Circle. — Let it be required to ascertain the moments 
and product of inertia of the area of fig. D 1 2 with reference to any 
coordinate axes u and v through the point 0. Let Ox and Oy be 
two axes with respect to which the moments and product of inertia 
of the area are known. Then lay off 

OX=I x , OY=I y , and YA=J xy , 

OX and OY along the positive x axis and YA in the positive or nega- 
tive y direction according as J X y is positive or negative. The circle 

* The invention of the inertia-circle is due to Prof. Culmann. For a 
full treatment see his ," Graphische Statik" or that of Miiller-Breslau, or 
a paper by Prof. L. J. Johnson in the Jour. Assoc. Eng. Soc, No. 5, Vol. 
XXVIII. 



SECOND MOMENTS OF AREAS. 



375 



whose centre is midway between X and Y (the point C) and passing 
through A is the inertia-circle for the area corresponding to the 
axes x and y. 




Fig. D 12. 
To determine I u and J uv . — From A draw a secant parallel to the 
u axis, thus determining a point B, and from B drop a perpendicular 
to the x axis, thus determining a point U. Then 

OU = I u and UB=J U v 
to the scale used in laying off OX, OF, and YA* J uv is positive or 

* Proof: In part, fig. D 1 3 is a reproduction of a portion of fig. D 1 2 (a) ; 
XA' is made equal to Y A, hence A'Bb 
is perpendicular to AB and to the it 
axis. The construction of the other 
dotted lines is obvious, they being 
either parallel or perpendicular to the 
u axis. 

Equation (1), art. D 11 can be 
written thus: 

Iu = (Jx cos a — J xy sin a) cos a 

+ (I V sin a— Jxy cos a) sin a, Fig. D 13. 

and this expression for I u can be readily evaluated from the figure. It 
is plain that since OX =I X and XA' =Jxy, 

I x cos a =0a, J X y sin a =ab, I x cos a —Jxy sin a —Ob, 
and {I x cos a— J xy sin a) cos a : =0c; 




376 



APPENDIX D. 



negative according as UB is along the positive or negative y direc- 
tion. If it should happen that Juv equals zero, the construction 
would simplify ; for in that case A coincides with Y, and hence XY 
is a diameter of the inertia-circle. 



EXAMPLES. 

i , art. D 1 1 by means of the inertia-circle. 



i. Solve ex. 

Solution: Lay off on the x axis (fig. D 14) OX, OY, and YA to 




Fig. D 14. 



Fig. D 15. 



represent I x , Iy, and J X y (scale 1 in. =30 in. 4 ). Then draw a circle 
with centre at C and radius equal to CA ; this is the inertia-circle 
corresponding to the axis Ox and Oy. Next draw a line through A 
parallel to the u axis, mark its intersection with the circle B, and 
drop the perpendicular BU ; then OU (1.2 in.) and UB (0.62 in.) 
represent the moment of inertia and product of inertia sought. 
<2. Solve ex. 2, art. D 11 by means of the inertia-circle. 

also that, since Oy =I y and YA =J X y, 

I y sin a = Yd, J xy cos a = Ye, I y sin a —J X y cos a=de =Bb, 
and {I y sin a —Jxy cos a) sin a=Bf =cU. 

Hence I u =Oc + cU =0U. q.e.d. 

Equation (2), art. D 11 can be written thus: 

Juv — (Ix cos a —Jxy sin a) sin a — (I y sin a —Jxy cos a) cos a, 
and this expression for J uv can be readily evaluated from the figure. As 
already explained, 

I x cos a —Jxy sin a =ob, 
hence (I x cos ct — J xy sin a) sin a =bc; 

and since, as explained, I y sin a—J xy cos a=Bb, 

(I x sin a— J xy sin a) cos a =bf. 
Hence J xy =bc — bf=fc=BU. q.e.d. 



SECOND MOMENTS OF AREAS. 377 

D 14. Principal Axes and Principal Moments of Inertia. — As the angle a 
(fig. D 12) increases from o to 360 , the point U moves along the x 
axis, its extreme positions being P x and P 2 , and for these two posi- 
tions UB equals zero. Hence 

(a) the maximum and minimum values of I u are given by OPi 

and 0P 2 ; 

(b) the corresponding inertia-axes are parallel to AP X and AP, 

(therefore rectangular) ; 

(c) the product of inertia with respect to those axes equals zero. 
If I x equals I y and J xy = o, the inertia-circle vanishes, and hence 

Iu is constant and Juv equals zero for all values of a. 

Definitions. — The two axes through a point with respect to 
which the moments of inertia of an area are greater and less than 
for any other axes through that point are called the principal axes of 
the figure at that point, and the corresponding moments of inertia 
and radii of gyration are called the principal moments of inertia and 
radii of gyration of the area at that point. We will denote these 
maximum and minimum moments of inertia and radii of gyration 
by 7 P I 2 , k v and k 2 respectively, and will mark the corresponding 
inertia-axes (1) and (2) (see fig. D 12). 

According to the above the product of inertia of an area with 
respect to the principal axes at a point equals zero. This proposi- 
tion leads at once to an algebraic method for finding principal axes 
and moments of inertia. Thus let a! denote the value of a which 
makes J uv zero, then (see eq. (2), art. D n) 

i(Ix-Iy) Sin 20t' + Jxy COS 2a' = 0, 

or tan 2a' = 2j xy /(Iy — Ix) (1) 

By means of this equation we may locate the principal axes at a 
point (i.e., determine a'), and then determine the principal moments 
by substituting the two values of a' given by that equation in eq. 
(1), art. D n. 

EXAMPLES. 

i. Determine the central principal axes of the angle section of 
fig. D 6 and the corresponding moments of inertia, I x , Iy, and J xv , 
being 45.37, 7.53, and —9.67 in. respectively. 

Solutions: (1) Graphical. Fig. D 15 is the inertia-circle for the 
area corresponding to the axis Ox and Oy (constructed as explained 
in the solution of ex. 1, art. D 13). 0(i) and 0(2), parallel to AP X 
and AP 2 , are the principal axes at 0, and 0P X and 0P 2 represent 
(by the scale used) the greater and lesser principal moments re- 
spectively. 



378 APPENDIX D. 

(2) Algebraic, Substituting in equation (1), we find that 

tan 2a / = 2( -9.67)7(7.53 -45.32) =0.5118, 

i.e., 2a' = 27° 6' or 207 6'; hence a' = 13° 2>2> or io 3° 33'- 

Substituting these two values successively in eq. (1), art. D 11, 
we find as the two values of I u 

l x =47.70 in. 4 and J 2 = 5.2oin. 4 

2. In fig. D 6 (6) let a, 6, and t equal 6, 3^, and f in. respectively; 
then 

I x = 2$.^2, Iy = g. 11, and Jxy= —11.48 in. 4 

Determine the central principal axes and the corresponding mo- 
ments of inertia. 

3. In fig. D6(a) let AB = BC = 3 in., AA' = CC =\ in.; then 
the centroid is 0.84 in. from AB and BC and I x =1^ = 1.24 in.* 
Determine the central principal axes and the corresponding radii 
of gyration. Ans. k 2 = o.$g in. 



INDEX. 



Acceleration. 172, 186 

angular, 205, 208 
resolution of, 1 88- 192 
Acceleration, time curves, 175 
Action and reaction, 3, 221 
Activity, 318 
Amplitude, 177 
Analysis, methods of, 2 
Angle of repose, 150 
Angular acceleration, 205, 208 
displacement, 204, 208 
impulse, 331 
momentum, 333, 334 
velocity, 204, 208 
Attraction, electric, 81-84 

gravitational, 74-81 
magnetic, 81-84 

Balancing, 279-283 
Blow, force of a, 343, 344 

Catenary, 109 
Centrifugal force, 261 
Centripetal force, 261 
Centre of gravity, defined, 54 

determination of, 55-59 
Centre of mass, 220, 221 

of percussion, 347, 348 

of stress, 88 
Centrode, 213 
Centroid, defined, 52, 60 

determination of, 53, 60-73 
Collision, 344-347 
Components of a force, 10-12, 14, 15 

Numbers 



Composition of couples, 49 
of forces, 22-46 
of forces, defined, 10 
of harmonic motions, 196- 
202 
Compression, 86 
Conservation of energy, 316 
Cords, flexible, 103-113 
Couple, defined, 9, 19 

graphical representation of, 20 
moment of, 19 
resolution of, 50 
Couples, theory of, 47-51 

D'Alembert's principle, 229 
Density, 75 

Dimensions of units, 357-362 
Displacement, 167, 177, 183 

angular, 204, 208 

defective, 298 
Dynamics, defined, 1 
Dyne, 223, 226 

Effective force, 228 
Efficiency, 317 

of tackle, 324-326 
of a mine-hoist, 326, 327 
Energy, 306-313 

conservation of, 316 
kinetic, 306-308 
mechanical, 312 
potential, 309-312 
principle of, for machines, 317 
Equilibrium, conditions of, 93, 95-101 
defined, 93 
refer to pages. 

379 



3 8o 



INDEX. 



Field, strength of, 76, 82 
Force at a distance, 3 
centrifugal, 261 
centripetal, 261 
components of, 10-12, 14, 15 
concentrated, 4 
contact, 3 
defined, 3 
distributed, 4 
effective, 228 
external, 93 

graphical representation of, 5 
internal, 93 
line of action of, 4 
moment of, 16, 17 
of a blow, 343 
parallelogram, 11 
polygon, 23 

resolution of, defined, 10 
systems, 9 

transmissibility of, 10 
triangle, 11 
units of, 6, 223 
Forces, composition of, 22—46 

composition of, defined, 10 
concurrent, defined, 9 
conservative, 311 
coplanar, defined, 9 
non-concurrent, defined, 9 
non-conservative, 311 
non-coplanar, defined, 9 
resultant of, defined, 10 
Frequency, 177 
Friction, 149-158, 247, 248 
angle of, 150 
belt, 157 
brake, 323 
circle, 154 

coefficient of, 150, 151, 247 
cone, 156 
journal, 285, 286 
kinetic, 247, 248 
laws of, 151, 247, 285 
pivot, 283 
rolling, 295-297 

Numbers 



Friction, static, defined, 149 
Funicular polygon, 26 

Geekilogram, 224-226 
Geepound, 224-226 
Graphical analysis, 2 
Gravitation constant, 74 
law of, 74 

Harmonic motion, composition of, 196- 
202 

resolution of, 198. 201 
simple, 177-180 
Hinge reactions, 275 
Hodograph, 185 
Hoop tension, 274 
Horse-power, 318 

Impact, 344-347 
Impulse, 329, 331 

angular, 331 
moment of, 331 
sudden, 343 
Inertia circle, 374-376 

ellipse, 374 
Instantaneous axis, 212 
centre, 213 
rotation, 213 

Jet, pressure due to a, 34.0, 341 

Kilogram, 6, 7 
Kinematics, 167 

defined, 1 
Kinetic energy, 306-308 
friction, 247, 248 
reactions, 239, 271, 292 
units, 223 
Kinetics, 217 

defined, 1 

Laws of motion, 221, 223 
Lag, 177 
Lead, 177 

refer to pages. 



INDEX. 



381 



Mass, defined, 6, 219 

moment of, 220, 249-257 
units of, 6, 223-226 
centre, 220, 221 
Mechanics, lefined, 1, 2 

subdivisions of, I 
Moment of a couple, 19 

of a force, 16, 17 

of a length, 61 

of a mass, 220, 249-257 

of a momentum, 333, 334 

of a volume, 61 

of a weight, 55 

of an area, 61, 363-378 

of an impulse, 331 

of inertia, 249-257, 363-378 

of inertia, experimental deter 

mination of, 255 
of inertia, principal, 377, 378 
Moments, principal of, 28, 32, 39, 44 
Momentum, 332-336 

angular, 333, 334 
moment of, 333, 334 
Motion, curvilinear, 183-202 
laws of, 221, 223 
non-uniform, 168, 172, 184 
of a particle, 221-228 
of a rigid body, 203-216, 233- 

297 
of a system of particles, 228- 

232 
plane, 207-216, 287-297 
rectilinear, 167-182 
relativity of, 192-196 
uniform, 168, 176, 184 
uniformly accelerated, 176 

Neutral axis, 89 

Pappus and Guldinius, theorem of, 72 
Parallelogram of forces, 1 1 
Parallelopiped of forces, 13 
Pendulum, ballistic, 347 
conical, 269 
Pendulums, 265-268, 269, 270 



Percussion, centre of, 347 

Period, 177 

Phase, 177 } 

Potential energy, 309-312 

Pound, 6, 7 

Poundal, 224-226 

Power, 318 

Principal axes, 257, 377 

Product of inertia, 257, 371-373 

Pulley, 113, 114, 325 

Radius of gyration, 250, 364 
principal, 377 

Rates, 351-356 

Repose, angle of, 150 

Resolution of acceleration) 188-192 
of a couple, 50 
of a force, defined, 10 
of harmonic motion, 198, 201 

Resultant of forces, defined, 10 

Rigid body, defined, 228 

Rolling resistance, 295-297 

Rotation, 203-207, 249-286 

Scalar, 349 

Shear, 86 

Space diagram, 5 

Space-time curve, 167, 183 

Speed-time curve, 185 

Statics, 3 

defined, 1 
Stress, 85-92 

centre of, 88 

diagram, 145 

intensity of, 86 

Tackle, 115, 116 

efficiency of, 324-326 
Tension, 86 
Torsion balance, 269 
Train- resistance, 321 
Translation, 203, 233-248 
Transmissibility of force, IO 
Triangle of forces, n 
Trusses, analysis of, 136-148 



Numbers refer to pages. 



382 



INDEX. 



Units, absolute, 7, 223 
derived, 357 
dimensions of, 357-362 
fundamental, 357 
gravitational, 6, 224 
kinetic, 223 

Varignon's theorem, 17 
Vectors, 349, 350 
Vector diagram, 5 
Velocity, 168, 184 



Velocity, angular, 204, 208 

resolution of, 188-192 
Velocity-time curve, 171 
Vibrations, 242-247 

Watt, 318 

Weight, apparent, 274 

defined, 8 
Work, 298-305 

and energy, 314, 328 

diagram, 299 



Numbers refer to pages. 



SHORT-TITLE CATALOGUE 

OF THE 

PUBLICATIONS 

OF 

JOHN WILEY & SONS, 

New York. 
London: CHAPMAN & HALL, Limited. 



ARRANGED UNDER SUBJECTS. 



Descriptive circulars sent on application. Books marked with an asterisk are 
sold at net prices only, a double asterisk (**) books sold under the rules of the 
American Publishers' Association at net prices subject to an extra charge for 
postage. All books are bound in cloth unless otherwise stated. 



AGRICULTURE. 



Armsby's Manual of Cattle-feeding i2mo, Si 75 

Principles of Animal Nutrition 8vo, 4 00 

Budd and Hansen's American Horticultural Manual: 

Part I. — Propagation, Culture, and Improvement i2mo, 1 50 

Part II. — Systematic Pomology i2mo, 1 50 

Downing's Fruits and Fruit-trees of America 8vo, 5 00 

Elliott's Engineering for Land Drainage i2mo, 1 50 

Practical Farm Drainage i2mo, 1 o 

Green's Principles of American Forestry. (Shortly.) 

Grotenfelt's Principles of Modern Dairy Practice. (Woll.) 12 mo, 20 

Kemp's Landscape Gardening i2mo, 2 50 

Maynard's Landscape Gardening as Applied to Home Decoration i2mo, 1 s 

Sanderson's Insects Injurious to Staple Crops i2mo, 1 5 

Insects Injurious to Garden Crops. (In preparation.) 
Insects Injuring Fruits. (In preparation.) 

Stockbridge's Rocks and Soils 8vo, 2 

Woll's Handbook for Farmers and Dairymen i6mo, 1 50 

ARCHITECTURE. 

Baldwin's Steam Heating for Buildings i2mo, 2 50 

Berg's Buildings and Structures of American Railroads 4to, 5 00 

Birkmire's Planning and Construction of American Theatres 8vo, 3 00 

Architectural Iron and Steel 8vo, 3 50 

Compound Riveted Girders as Applied in»Buildings 8vo, 2 00 

Planning and Construction of High Office Buildings 8vo, 3 50 

Skeleton Construction in Buildings 8vo, 3 00 

Briggs's Modern American School Buildings 8vo, 4 00 

Carpenter's Heating and Ventilating of Buildings 8vo, 4 00 

Freitag's Architectural Engineering. 2d Edition, Rewritten 8vo, 3 50 

Fireproofing of Steel Buildings 8vo, 2 50 

French and Ives's Stereotomy 8vo, 2 50 

Gerhard's Guide to Sanitary House-inspection i6mo, 1 00 

Theatre Fires and Panics i2mo, 1 50 

1 



Hatfield's American House Carpenter 8vo, 

Holly's Carpenters' and Joiners' Handbook " i8mo, 

Johnson's Statics by Algebraic and Graphic Methods . .8vo, 

Kidder's Architect's and Builder's Pocket-book i6mo, morocco, 

Merrill's Stones for Building and Decoration 8vo, 

Monckton's Stair-building . 4 to, 

Patton's Practical Treatise on Foundations 8vo, 

Siebert and Biggin's Modern Stone-cutting and Masonry 8vo, 

Snow's Principal Species of Wood 8vo, 

Sondericker's Graphic Statics with. Applications to Trusses, Beams, and Arches. 
(Shortly.) 

Wait's Engineering and Architectural Jurisprudence 8vo, 

Sheep, 
Law of Operations Preliminary to Construction in Engineering and Archi- 
tecture .8vo, 

Sheep, 

Law of Contracts 8vo, 

Woodbury's Fire Protection of Mills 8vo, 

Worcester and Atkinson's Small Hospitals, Establishment and Maintenance, 
Suggestions^for Hospital Architecture, with Plans for a Small Hospital. 

i2mo, 
The World's Columbian Exposition of 1893 Large 410, 

ARMY AND. NAVY. 

Bernadou's Smokeless Powder, Nitro-cellulose, and the Theory of the Cellulose 
Molecule i2mo, 

* Bruff's Text-book Ordnance and Gunnery 8vo, 

Chase's Screw Propellers and Marine Propulsion 8vo, 

Craig's Azimuth 4to, 

Crehore and Squire's Polarizing Photo-chronograph 8vo, 

Cronkhite's Gunnery for Non-commissioned Officers 24mo. morocco, 

* Davis's Elements of Law 8vo, 

* Treatise on the Military Law of United States 8vo, 

* Sheep 

De Brack's Cavalry Outpost Duties. (Carr.) 24100, morocco, 

Dietz's Soldier's First Aid Handbook i6mo, morocco, 

* Dredge's Modern French Artillery 4to, half morocco, 

Durand's Resistance and Propulsion of Ships 8vo, 

* Dyer's Handbook of Light Artillery i2mo, 

Eissler's Modern High Explosives 8vo, 

* Fiebeger's Text-book on Field Fortification Small 8vo, 

Hamilton's The Gunner's Catechism i8mo, 

* Hoff 's Elementary Naval Tactics 8vo, 

Ingalls's Handbook of Problems in Direct Fire 8vo, 

* Ballistic Tables, 8vo> 

* Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and II . . 8vo, each, 

* Mahan's Permanent Fortifications. (Mercur.) 8vo, half morocco, 

Manual for Courts-martial i6mo morocco, 

* Mercur's Attack of Fortified Places nmo, 

* Elements of the Art of War 8vo, 

MetcalfVCost of Manufactures — And the Administration of Workshops, Public 

and Private 8vo, 

* Ordnance and Gunnery nmo, 

Murray's Infantry Drill Regulations i8mo, paper, 

* Phelps's Practical Marine Surveying 8vo, 

Powell's Army Officer's Examiner nmo, 

Sharpe's Art of Subsisting Armies in War i8mo, morocco, 

<> 



5 


OO 




75 


2 


00 


4 


00 


5 


00 


4 


00 


5 


00 


1 


50 


3 


50 


6 


00 


6 


50 


5 


00 


5 


50 


3 


00 


2 


50 


I 


25 


I 


00 



2 


50 


6 


00 


3 


00 


3 


5o 


3 


00 


2 


00 


2 


50 


7 


00 


7 


50 


2 


00 


1 


25 


15 


00 


5 


00 


3 


00 


4 


00 


2 


00 


1 


00 


1 


50 


4 


00 


1 


50 


6 


00 


7 


50 


1 


SO 


2 


00 


4 


00 


5 


00 


5 


00 




10 


2 


50 


4 


00 


I 


So 



I 


50 


3 


oo 


i 


00 


2 


oo 


3 


oo 


3 


oo 


I 


50 


I 


50 



* Walke's Lectures on Explosives 8vo, 4 00 

* Wheeler's Siege Operations and Military Mining . . . .* 8vo, 2 00 

Winthrop's Abridgment of Military Law i2mo, 2 50 

Woodhull's Notes on Military Hygiene i6mo» 1 50 

Young's Simple Elements of Navigation i6mo, morocco, 1 00 

Second Edition, Enlarged and Revised i6mo, morocco 2 00 



ASSAYING. 

Fletcher's Practical Instructions in Quantitative Assaying with the Blowpipe. 

i2mo, morocco, 

Furman's Manual of Practical Assaying 8vo, 

Miller's Manual of Assaying i2mo, 

O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 

Ricketts and Miller's Notes on Assaying 8vo, 

Ulke's Modern Electrolytic Copper Refining 8vo, 

Wilson's Cyanide Processes i2mo, 

Chlorination Process i2mo . 



ASTRONOMY. 

Comstock's Field Astronomy for Engineers 8vo, 2 50 

raig's Azimuth 4to, 3 50 

Doolittle's Treatise on Practical Astronomy 8vo, 4 00 

Gore's Elements of Geodesy 8vo, 2 50 

Hayford's Text-book of Geodetic Astronomy 8vo, 3 00 

Merriman's Elements of Precise Surveying and Geodesy 8vo, 2 50 

* Michie and Harlow's Practical Astronomy 8vo, 3 00 

* White's Elements of Theoretical and Descriptive Astronomy i2mo, 2 00 



BOTANY. 

Davenport's Statistical Methods, with Special Reference to Biological Variation. 

i6mo, morocco, 1 23 

Thome and Bennett's Structural and Physiological Botany i6mo, 2 25 

Westermaier's Compendium of General Botany. (Schneider.) 8vo, 2 00 



CHEMISTRY. 

Adriance's Laboratory Calculations and Specific Gravity Tables i2mo, 1 25 

Allen's Tables for Iron Analysis 8vo, 3 00 

Arnold's Compendium of Chemistry. (Mandel.) (In preparation.) 

Austen's Notes for Chemical Students i2mo, 1 50 

Bernadou's Smokeless Powder. — Nitro-cellulose, and Theory of the Cellulose 

Molecule i2mo, 

Bolton's Quantitative Analysis 8vo, 

* Browning's Introduction to the Rarer Elements 8vo, 

Brush and Penfield's Manual of Determinative Mineralogy 8vo, 

Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.) . . . .8vo 

Cohn's Indicators and Test-papers i2mo, 

Tests and Reagents 8vo, 

Copeland's Manual of Bacteriology. (In preparation.) 

Craft's Short Course in Qualitative Chemical Analysis. (Schaeffer.). . . . i2mo, 

Drechsel's Chemical Reactions. (Merrill.) i2mo, 

Duhem's Thermodynamics and Chemistry. (Burgess.) (Shortly.) 

Eissler's Modern High Explosives 8vo, 4 00 



2 


50 


I 


50 


I 


50 


4 


00 


3 


00 


2 


00 


3 


00 


2 


00 


1 


25 



Effront's Enzymes and their Applications. (Prescott.) 8vo, 3 00 

Erdmann's Introduction to Chemical Preparations. (Dunlap.) i2mo, 1 25 

Fletcher's Practical Instructions in Quantitative Assaying with the Blowpipe. 

i2mo, morocco, 1 50 

Fowler's Sewage Works Analyses i2mo, 2 00 

Fresenius's Manual of Qualitative Chemical Analysis. (Wells.) 8vo, 5 00 

Manual of Qualitative Chemical Analysis. Parti. Descriptive. (Wells.) 

8vo, 3 00 

System of Instruction in Quantitative Chemical Analysis. (Cohn.) 
2 vols. (Shortly.) 

Fuertes's Water and Public Health i2mo, 

Furman's Manual of Practical Assaying 8vo, 

Gill's Gas and Fuel Analysis for Engineers i2mo, 

Grotenfelt's Principles of Modern Dairy Practice. (Woll.) nmo, 

Hammarsten's Text-book of Physiological Chemistry. (Mandel.) 8vo, 

Helm's Principles of Mathematical Chemistry. (Morgan.) i2mo. 

Hinds's Inorganic Chemistry . . . -. 8vo, 

* Laboratory Manual for Students i2mo, 

Holleman's Text-book of Inorganic Chemistry. (Cooper.) 8vo, 

Text-book of Organic Chemistry. (Walker and Mott.) 8vo, 

Hopkins's Oil-chemists' Handbook. 8vo, 

Jackson's Directions for Laboratory Work in Physiological Chemistry. .8vo, 

Keep's Cast Iron 8vo, 

Ladd's Manual of Quantitative Chemical Analysis i2mo. 

Landauer's Spectrum Analysis. (Tingle.) 8vo, 

Lassar-Cohn's Practical Urinary Analysis. (Lorenz.) 12 mo. 

Leach's The Inspection and Analysis of Food with Special Reference to State 

Control. (In preparation.) 
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) i2mo, 

Mandel's Handbook for Bio-chemical Laboratory i2mo, 

Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.) 

3d Edition, Rewritten 8vo, 

Examination of Water. (Chemical and Bacteriological.) i2mo, 

Meyer's Determination of Radicles in Carbon Compounds. (Tingle.). . i2mo, 

Miller's Manual of Assaying i2mo, 

Mixter's Elementary Text-book of Chemistry i2mo, 

Morgan's Outline of Theory of Solution and its Results nmo, 

Elements of Physical Chemistry i2mo. 

Nichols's Water-supply. (Considered mainly from a Chemical and Sanitary 

Standpoint, 1883.) 8vo, 

O'Brine's Laboratory Guide in Chemical Analysis 8vo, 

O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 

Ost and Kolbeck's Text-book of Chemical Technology. (Lorenz — Bozart.) 

(In preparation.) 

* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo, paper, 50 
Pictet's The Alkaloids and their Chemical Constitution. (Biddle.) (In 
preparation.) 

Pinner's Introduction to Organic Chemistry. (Austen.) i2mo, 1 50 

Poole's Calorific Power of Fuels 8vo> 3 00 

* Reisig's Guide to Piece-dyeing 8vo, 25 00 

Richards and Woodman's Air .Water , and Food from a Sanitary'Standpoint . 8vo, 2 00 
Richards's Cost of Living as Modified by Sanitary Science i2mo, 1 00 

Cost of Food a Study in Dietaries i2mo, 1 00 

* Richards and Williams's The Dietary Computer 8vo, 1 50 

Ricketts and Russell's Skeleton Notes upon Inorganic Chemistry. (Part I. — 

Non-metallic Elements.) 8vo, morocco, 75 

Ricketts and Miller's Notes on Assaying 8vo, 3 00 

4 



I 


50 


3 


OO 


1 


25 


2 


OO 


4 


OO 


1 


50 


3 


OO 




75 


2 


50 


2 


50 


3 


OO 


I 


OO 


2 


50 


1 


00 


3 


OO 


1 


OO 


1 


OO 


1 


50 


4 


OO 


1 


25 


1 


OO 


1 


OO 


1 


50 


1 


OO 


2 


OO 


2 


50 


2 


OO 


2 


OO 



d eal's Sewage and the Bacterial Purification of Sewage 8vo, 3 50 

Ruddiman's Incompatibilities in Prescriptions 8vo, 2 00 

Schimpf's Text-book of Volumetric Analysis i2mo, 2 50 

Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, morocco, 3 00 

Handbook for Sugar Manufacturers and their Chemists. . i6mo, morocco, 2 00 

Stockbridge's Rocks and Soils 8vo, 2 50 

* Tillman's Elementary Lessons in Heat 8vo, 1 50 

* Descriptive General Chemistry 8vo 3 00 

Treadwell's Qualitative Analysis. (Hall.) 8vo, 3 00 

Turneaure and Russell's Public Water-supplies 8vo, 5 00 

Van Deventer's Physical Chemistry for Beginners. (Boltwood.) i2mo, 1 50 

* Walke's Lectures on Explosives 8vo, 4 00 

Wells's Laboratory Guide in Qualitative Chemical Analysis 8vo, 1 50 

Short Course in Inorganic Qualitative Chemical Analysis for Engineering 

Students i2mo, 1 50 

Whipple's Microscopy of Drinking-water 8vo, 3 50 

Wiechmann's Sugar Analysis Small 8vo. 2 50 

Wilson's Cyanide Processes i2mo, 1 50 

Chlorination Process i2mo, 1 50 

Wulling's Elementary Course in Inorganic Pharmaceutical and Medical Chem- 
istry i2mo, 2 00 

CIVIL ENGINEERING. 

BRIDGES AND ROOFS. HYDRAULICS. MATERIALS OF ENGINEERING. 
RAILWAY ENGINEERING. 

Baker's Engineers' Surveying Instruments i2mo, 3 00 

Bixby's Graphical Computing Table Paper, io£X 24^ inches 25 

** Burr's Ancient and Modern Engineering and the Isthmian Canal. (Postage 

27 cents additional.) 8vo, nt 3 50 

Comstock's Field Astronomy for Engineers 8vo, 2 50 

Davis's Elevation and Stadia Tables 8vo, 1 00 

Elliott's Engineering for Land Drainage. i2mo, 1 50 

Practical Farm Drainage i2mo, 1 00 

FolwelTs Sewerage. (Designing and Maintenance.) 8vo, 3 00 

Freitag's Architectural Engineering. 2d Edition, Rewritten 8vo, 3 50 

French and Ives's Stereotomy 8vo, 2 50 

Goodhue's Municipal Improvements i2mo, 1 75 

Goodrich's Economic Disposal of Towns' Refuse 8vo, 3 50 

Gore's Elements of Geodesy 8vo, 2 50 

Hayford's Text-book of Geodetic Astronomy 8vo, 3 00 

Howe's Retaining Walls for Earth i2mo, 1 25 

Johnson's Theory and Practice of Surveying Small 8vo, 4 00 

Statics by Algebraic and Graphic Methods 8vo, 2 00 

Kiersted's Sewage Disposal i2mo, 1 25 

Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) i2mo, 200 

Mahan's Treatise on Civil Engineering. (1873.) (Wood.) 8vo 5 00 

* Descriptive Geometry 8vo, 1 50 

Merriman's Elements of Precise Surveying and Geodesy 8vo, 2 50 

Elements of Sanitary Engineering 8vo, 2 00 

Merriman and Brooks's Handbook for Surveyors i6mo, morocco, 2 00 

Nugent's Plane Surveying 8vo, 3 50 

Ogden's Sewer Design i2mo, 2 00 

Patton's Treatise on Civil Engineering 8vo, half leather, 7 50 

Reed's Topographical Drawing and Sketching 4to, 5 00 

Rideal'sJSewage and the Bacterial Purification of Sewage 8vo, 3 50 

Siebert and Biggin's Modern Stone-cutting and Masonry 8vo, 1 so 

Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 2 so 

5 



Sondericker's Graphic Statics, wxtn i-pplications to Trusses, Beams, ana 
Arches. (Shortly.) 

* Trautwine's Civil Engineer's Pocket-book i6mo, morocco, 5 00 

Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 

Sheep, 6 50 
Law of Operations Preliminary to Construction in Engineering and Archi- 
tecture, 8vo, 5"oo 

Sheep, 5 5o 

Law of Contracts 8vo, 3 00 

Warren's Stereotomy — Problems in Stone-cutting 8vo, 2 50 

Webb's Problems in the U»e and Adjustment of Engineering Instruments. 

i6mo, morocco, 1 25 

* Wheeler's Elementary Course of Civil Engineering 8vo, 4 00 

Wilson's Topographic Surveying 8vo, 3^50 



BRIDGES AND ROOFS. 

Boiler's Practical Treatise on the Construction of Iron Highway Bridges. .8vo, 2 00 

* Thames River Bridge 4to, paper, 5 00 

Burr's Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and 

Suspension Bridges 8vo, 3 50 

Du Bois's Mechanics of Engineering. Vol. II Small 4to, 10 00 

Foster's Treatise on Wooden Trestle Bridges 4to, 5 00 

Fowler's Coffer-dam Process for Piers 8vo, 2 50 

Greene's Roof Trusses 8vo, 1 25 

Bridge Trusses 8vo, 2 50 

Arches in Wood, Iron, and Stone 8vo, 2 50 

Howe's Treatise on Arches 8vo 4 00 

Design of Simple Roof-trusses in Wood and Steel 8vo, 2 00 

Johnson, Bryan, and Turneaure's Theory and Practice in the Designing of 

Modern Framed Structures Small 4to, 10 00 

Merriman and Jacoby's Text-book on Roofs and Bridges: 

Part I. — Stresses in Simple Trusses 8vo, 2 50 

Part II. — Graphic Statics 8vo, 2 50 

Part III. — Bridge Design. 4th Edition, Rewritten 8vo, 2 50 

Part IV. — Higher Structures 8vo, 2 50 

Morison's Memphis Bridge 4to, 10 00 

Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . . i6mo, morocco, 3 00 

Specifications for Steel Bridges i2mo, 1 25 

Wood's Treatise on the Theory of the Construction of Bridges and Roofs. 8vo, 2 00 
Wright's Designing of Draw-spans: 

Part I. — Plate-girder Draws 8vo, 2 50 

Part II. — Riveted-truss and Pin-connected Long-span Draws 8vo, 2 50 

Two parts in one volume 8vo, 3 50 



HYDRAULICS. 

Bazin's Experiments upon the Contraction of the Liquid Vein Issuing from an 

Orifice. (Trautwine.) 8vo, 2 00 

Bovey's Treatise on Hydraulics 8vo, 5 00 

Church's Mechanics of Engineering 8vo, 6 00 

Diagrams of Mean Velocity of Water in Open Channels paper, 1 50 

Coffin's Graphical Solution of Hydraulic Problems i6mo, morocco, 2 50 

Flather's Dynamometers, and the Measurement of Power nmo, 3 00 

Folwell's Water-supply Engineering 8vo, 4 00 

Frizell's Water-power 8vo, 5 00 

6 



Fuertes's Water and Public Health izmo, I 50 

Water-filtration Works i2mo, 2 50 

Ganguillet and Kutter's General Formula for the Uniform Flow of Water in 

Rivers and Other Channels. (Hering and Trau twine.) 8vo, 4 00 

Hazen's Filtration of Public Water-supply 8vo, 3 00 

Hazlehurst's Towers and Tanks for Water- works 8vo, 2 50 

Herschel's 115 Experiments on the Carrying Capacity of Large, Riveted, Metal 

Conduits 8vo, 2 00 

Mason's Water-supply. (Considered Principally from a Sanitary Stand- 
point.) 3d Edition, Rewritten 8vo, 4 00 

Merriman's Treatise on Hydraulics. 9th Edition, Rewritten 8vo, 5 00 

* Michie's Elements of Analytical Mechanics 8vo, 4 00 

Schuyler's Reservoirs for Irrigation, Water-power, and Domestic Water- 
supply Large 8vo, s 00 

** Thomas and Watt's Improvement of Riyers. (Post., 44 c. additional), 4to, 6 00 

Turneaure and Russell's Public Water-supplies 8vo. 5 00 

Wegmann's Desien and Construction of Dams 4to, 5 00 

Water-suoolv of the City of New York from 1658 to 1895 4to, 10 00 

Weisbach's Hydraulics and Hydraulic Motors. (Du Bois.) 8vo, 5 00 

Wilson's Manual of Irrigation Engineering Small 8vo, 4 00 

Wolff's Windmill as a Prime Mover 8VO, 1 3 00 

Wood's Turbines 8vo, 2 50 

Elements of Analytical Mechanics 8vo, 3 00 



MATERIALS OF ENGINEERING. 

Baker's Treatise on Masonry Construction 8vo, 

Roads and Pavements 8vo, 

Black's United States Public Works Oblong 4to, 

Bovey's Strength of Materials and Theory of Structures 8vo, 

Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edi- 
tion, Rewritten 8vo, 

Byrne's Highway Construction 8vo, 

Inspection of the Materials and Workmanship Employed in Construction. 

i6mo, 

Church's Mechanics of Engineering. 8vo, 

Du Bois's Mechanics of Engineering. Vol. I Small 4to, 

Johnson's Materials of Construction Large 8vo, 

Keep's Cast Iron 8vo, 

Lanza's Applied Mechanics 8vo, 

Martens's Handbook on Testing Materials. (Henning.) 2 vols 8vo, 

Merrill's Stones for Building and Decoration 8vo, 

Merriman's Text-book on the Mechanics of Materials 8vo, 

Strength of Materials nmo, 

Metcalf's Stefel. A Manual for Steel-users nmo, 

Patton's Practical Treatise on Foundations 8vo, 

Rockwell's Roads and Pavements in France i2tno, 

Smith's Wire : Its Use and Manufacture Small 4to, 

Materials of Machines 1 2mo, 

Snow's Principal Species of Wood 8vo, 

Spalding's Hydraulic Cement nmo, 

Text-book on Roads and Pavements nmo, 

Thurston's Materials of Engineering. 3 Parts 8vo, 

Part I. — Non-metallic Materials of Engineering and Metallurgy 8vo, 

Part H. — Iron and Steel 8vo, 

Part HI. — A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents 8vo, 2^50 

7 



5 


00 


5 


00 


5 


00 


7 


50 


7 


50 


5 


00 


3 


00 


6 


00 


7 


50 


6 


00 


2 


50 


7 


50 


7 


50 


5 


00 


4 00 


1 


00 


2 


00 


5 


00 


1 


25 


3 


00 


1 


00 


3 


50 


2 


00 


2 


00 


8 


00 


2 


00 


3 


50 



Thurston's Text-book of the Materials of Construction 8vo, 5 00 

Tillson's Street Pavements and Paving Materials 8vo, 4 00 

WaddelTs De Pontibus. (A Pocket-book for Bridge Engineers.)- . i6mo, mor., 3 00 

Specifications for Steel Bridges nmo, 1 25 

Wood's Treatise on the Resistance of Materials, and an Appendix on the Pres- 
ervation of Timber , 8vo, 2 00 

Elements of Analytical Mechanics 8vo, 3 00 



RAILWAY ENGINEERING. 

Andrews's Handbook for Street Railway Engineers. 3X5 inches, morocco, 1 25 

Berg's Buildings and Structures of American Railroads 4to, 5 00 

Brooks's Handbook of Street Railroad Location i6mo. morocco, 1 50 

Butts's Civil Engineer's Field-book i6mo, morocco, 2 50 

Crandall's Transition Curve i6mo, morocco, 1 50 

Railway and Other Earthwork Tables , 8vo, 1 50 

Dawson's "Engineering" and Electric Traction Pocket-book. i6mo, morocco, 4 00 

Dredge's History of the Pennsylvania Railroad: (1879) Paper, 5 00 

* Drinker's Tunneling, Explosive Compounds, and Rock Drills, 4to, half mor., 25 00 

Fisher's Table of Cubic Yards Cardboard, 25 

Godwin's Railroad Engineers' Field-book and Explorers' Guide i6mo, mor., 2 50 

Howard's Transition Curve Field-book i6mo morocco 1 50 

Hudson's Tables for Calculating the Cubic Contents of Excavations and Em- 
bankments 8vo, 1 00 

Molitor and Beard's Manual for Resident Engineers. i6mo, 1 00 

Nagle's Field Manual for Railroad Engineers . i6mo, morocco. 3 00 

Philbrick's Field Manual for Engineers i6mo, morocco, 3 00 

Pratt and Alden's Street-railway Road-bed : 8vo, 2 00 

Searles's Field Engineering i6mo, morocco, 3 00 

Railroad Spiral i6mo, morocco 1 50 

Taylor's Prismoidal Formulae and Earthwork 8vo, 1 50 

* Trautwine's Method of Calculating the Cubic Contents of Excavations and 

Embankments by the Aid of Diagrams 8vo, 2 00 

he Field Practice of [Laying Out Circular Curves for Railroads. 

i2mo, morocco, 2 50 

* Cross-section Sheet Paper, 25 

Webb's Railroad Construction. 2d Edition, Rewritten i6mo. morocco, 5 00 

Wellington's Economic Theory of the Location of Railways. .... .Small 8vo, 5 00 



DRAWING. 

Barr's Kinematics of Machinery 8vo, 

* Bartlett's Mechanical Drawing 8vo, 

Coolidge's Manual of Drawing 8vo, paper, 

Durley's Kinematics of Machines 8vo, 

Hill's Text-book on Shades and Shadows, and Perspective 8vo, 

Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 

Part II. — Form, Strength, and Proportions of Parts 8vo, 

MacCord's Elements of Descriptive Geometry 8vo, 

Kinematics; or, Practical Mechanism 8vo, 

Mechanical Drawing 4to, 

Velocity Diagrams 8vo, 

* Mahan's Descriptive Geometry and Stone-cutting 8vo, 

Industrial Drawing. (Thompson.) 8vo, 

Reed's Topographical Drawing and Sketching 4to, 



2 


50 


3 


00 


1 


00 


4 


00 


2 


00 


I 


50 


3 


00 


3 


00 


5 


00 


4 


00 


1 


50 


1 


50 


3 


50 


5 


00 



2 


00 


3 


00 


3 


00 


2 


50 


I 


00 


I 


25 


I 


50 


I 


00 


I 


25 




75 


3 


50 


3 


00 


7 


50 


2 


50 


5 


00 


2 


00 


3 


50 


2 


50 



Reid's Course in Mechanical Drawing 8vo, 

Text-book of Mechanical Drawing and Elementary Machine Design. .8vo, 

Robinson's Principles of Mechanism 8vo, 

Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 

Warren's Elements of Plane and Solid Free-hand Geometrical Drawing. . i2mo, 

Drafting Instruments and Operations i2mo, 

JUanual of Elementary Projection Drawing i2mo, 

Manual of Elementary Eroblems in the Linear Perspective of Form and 

Shadow i2mo, 

Plane Problems in Elementary Geometry i2mo, 

Primary Geometry i2mo, 

Elements of Descriptive Geometry, Shadows, and.Perspective . 8vo, 

General Problems of Shades and Shadows 8vo, 

Elements of Machine Construction and Drawing 8vo, 

Problems. Theorems, and Examples in Descriptive Geometrv 8vo, 

Weisbach's Kinematics and the Power of Transmission. (Hermann an<* 

Klein.) 8vo, 

Whelpley's Practical Instruction in the Art of Letter Engraving i2mo, 

Wilson's Topographic Surveying 8vo, 

Free-hand Perspective 8vo, 

Free-hand Lettering. (In preparation.) 
Woolf's Elementary Course in Descriptive Geometry Large 8vo, • 3 00 



'ELECTRICITY AND PHYSICS. 

Anthony and Brackett's Text-book of Physics. (Magie.) ... .Small 8vo, 3 00 

Anthony's Lecture-notes on the Theory of Electrical Measurements i2mo, 1 00 

Benjamin'sIHistory of Electricity 8vo, 3 00 

Voltaic Cell 8vo, 3 00 

Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.). .8vo, 3 00 

Crehore and Sauier's Polarizing Photo-chronograph 8vo, 3 00 

Dawson's "Engineering" and Electric Traction Pocket-book. . iomo, morocco, 4 00 

Flather's Dvnamometers, and the Measurement of Power nmo, 3 00 

Gilbert's De Magnete. (Mottelay.) 8vo, 2 50 

Holman's Precision of Measurements 8vo, 2 00 

Telescopic Mirror-scale Method, Adjustments, and Tests Large 8vo 75 

Lanaauer's Spectrum Analysis. (Tingle.) 8vo, 3 00 

Le Chatelier's High-temperature Measurements. (Boudouard — Burgess. )i2mo, 3 00 

Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) i2mo, 1 00 

* Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and 11. 8vo, each,5 6»oo 

* Michie. Elements of Wave Motion Relating tolSound'and Light Svo,^ 00 

Niaudet's Elementary Treatise on Electric Batteries. (FishpacK. ) i2mo, 2 50 

* Parshall and Hobart's Electric Generators Small 4to. half morocco, 10 00 

* Rosenberg's Electrical Engineering. (Haldane Gee — Kinzbrunner.). . . .8vo, 1 50 
Ryan, Norris, and Hoxie's Electrical Machinery. (In preparatior -• 

Thurston's Stationary Steam-engines 8vo, 2 50 

* Tillman's Elementary Lessons in Heat 8vo, 1 50 

Tory and Pitcher's Manual of Laboratory Physics Small 8vo, 2 00 

Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 



LAW. 

♦^Davis's Elements of Law 8vo, 2 50 

* Treatise on the Military Law of United States 8vo, 7 00 

* Sheep, 7 50 
Manual for Courts-martial i6mo, morocco, 1 50 



Wait's Engineering and Architectural Jurisprudence 8vo, 6 oo 

Sheep, 6 50 
Law of Operations Preliminary to Construction in Engineering'and Archi- 
tecture 8vo, 5 00 

Sheep, 5 50 

Law of Contracts 8vo, 3 00 

Winthrop's Abridgment of Military Law i2mo, ** 50 



MANUFACTURES. 

Bernadou's Smokeless Powder — Nitro-cellulose and Theory of the Cellulose 

Molecule i2mo, 2 50 

Bolland's Iron Founder i2mo, 2 50 

" The Iron Founder," Supplement i2mo, 2 50 

Encyclopedia of Founding and Dictionary of^Foundry Terms Used_m the 

Practice of Moulding i2mo, 3 00 

Eissler's Modern High Explosives 8vo, 4 00 

Effront's Enzymes and their Applications. (Prescott. ) 8vo, 3 00 

Fitzgerald's Boston Machinist i8mo, 1 00 

Ford's Boiler Making for Boiler Makers i8mo, 1 00 

Hopkins's Oil-chemists' Handbook 8vo, 3 00 

Keep's Cast Iron 8vo, 2 50 

Leach's The Inspection and Analysis of Food with SpeciallReference to State 

Control. (In preparation.) 

Metcalf's Steel. A Manual for Steel-users i2mo, 2 00 

Metcalfe's Cost of Manufactures — And the Administration of Workshops, 

Public and Private 8vo, 5 00 

Meyer's Modern Locomotive Construction 4to, 10 00 

* Reisig's Guide to Piece-dyeing 8vo, 25 00 

Smith's Press-working of Metals 8vo, 3 00 

Wire : Its Use and Manufacture Small 4to, 3 00 

Spalding's Hydraulic Cement i2mo, 2 00 

Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, morocco, 3 00 

Handbook tor sugar Manufacturers ana their Chemists.. . i6mo, morocco, 2 00 
Thurston's Manual of Steam-boilers, their Designs, Construction and Opera- 
tion 8vo, s 00 

* Walke's Lectures on Explosives 8vo, 4 00 

West's American Foundry Practice i2mo, 2 50 

Moulder's Text-book i2mo , 2 50 

Wiechmann's Sugar Analysis Small 8vo, 2 50 

Wolff's Windmill as a Prime Mover 8vo, 3 00 

Woodbury's Fire Protection of Mills 8vo, 2 50 



MATHEMATICS. 

Baker's Elliptic Functions 8vo, 1 50 

* Bass's Elements of Differential Calculus i2mo, 4 00 

Briggs's Elements of Plane Analytic Geometry i2mo, 1 00 

Chapman's Elementary Course in Theory of Equations i2mo, 1 50 

Compton's Manual of Logarithmic Computations i2mo, 1 50 

Davis's Introduction to the Logic of Algebra. 8vo, 1 50 

* Dickson's College Algebra Large i2mo, 1 50 

* Introduction to the Theory of Algebraic Equations LargeIi2mo, 1 25 

Halsted's Elements of Geometry 8vo, 1 75 

Elementary Synthetic Geometry 8vo. 1 50 

10 



* Johnson's Three-place Logarithmic Tables : Vest-pocket size paper, 

ioo copies for 

* Mounted on heavy cardboard, 8 X 10 inches, 

10 copies for 

Elementary Treatise on the Integral Calculus Small 8vo, 

Curve Tracing in Cartesian Co-ordinates i2mo, 

Treatise on Ordinary and Partial Differential Equations Small 8vo, 

Theory of Errors and the Method of Least Squares i2mo, 

* Theoretical Mechanics i2mo, 

Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) i2mo, 

* Ludlow and Bass. Elements of Trigonometry and Logarithmic and Other 

Tables 8vo, 

Trigonometry and Tables published separately Each, 

Maurer's Technical Mechanics. (In preparation.) 

Merriman and Woodward's Higher Mathematics 8vo, 

Merriman's Method of Least Squares 8vo, 

Rice and Johnson's Elementary Treatise on the Differential Calculus. Sm., 8vo, 

Differential and Integral Calculus. 2 vols, in one Gmall 8vo, 

Wood's Elements of Co-ordinate Geometry 8vo, 

Trigonometry: Analytical, Plane, and Spherical i2mo, 

MECHANICAL ENGINEERING. 

MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS. 

Baldwin's Steam Heating for Buildings i2mo, 

Barr's Kinematics of Machinery 8vo, 

* Bartlett's Mechanical Drawing 8vo, 

Benjamin's Wrinkles and Recipes i2mo, 

Carpenter's Experimental Engineering 8vo, 

Heating and Ventilating Buildings 8vo , 

Clerk's Gas and Oil Engine Small 8vo, 

Coolidge's Manual of Drawing 8vo, paper, 

Cromwell's Treatise on Toothed Gearing i2mo, 

Treatise on Belts and Pulleys i2mo, 

Durley's Kinematics of Machines 8vo, 

Flather's Dynamometers and the Measurement of Power nmo, 

Rope Driving i2mo, 

Gill's Gas and Fuel Analysis for Engineers nmo, 

Hall's Car Lubrication i2mo, 

Hutton's The Gas Engine. (In preparation.) 
Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 

Part H. — Form, Strength, and Proportions of Parts 8vo, 

Kent's Mechanical Engineer's Pocket-book i6mo, morocco, 

Kerr's Power and Power Transmission 8vo, 

MacCord's Kinematics; or, Practical Mechanism 8vo, 

Mechanical Drawing 4to, 

Velocity Diagrams 8vo, 

Mahan's Industrial Drawing. (Thompson.) 8vo, 

Poole's Calorific Power of Fuels 8vo, 

Reid's Course in Mechanical Drawing 8vo. 

Text-book of Mechanical Drawing and Elementary Machine Design. .8vo, 

Richards's Compressed Air i2mo, 

Robinson's Principles of Mechanism 8vo, 

Smith's Press-working of Metals - ,8vo 

Thurston's Treatise on Friction and Lost Work in Machinery and Mil 
Work 8vo, 

Animal as a Machine and Prime Motor, and the Laws of Energetics . i2mo, 

11 





15 


5 


00 




25 


2 


00 


I 


50 


I 


00 


3 


50 


I 


50 


3 


oo 


2 


00 


3 


oo 


2 


00 


5 


00 


2 


00 


3 


oo 


2 


50 


2 


00 


I 


00 



2 


50 


2 


50 


3 


00 


2 


00 


6 


00 


4 


00 


4 


00 


I 


00 


I 


50 


I 


50 


4 


00 


3 


00 


2 


00 


I 


25 


I 


00 


! 


50 


3 


00 


5 


00 


2 


00 


5 


00 


4 


00 


I 


50 


3 


50 


3 


00 


2 


00 


3 


00 


I 


50 


3 


00 


3 


oo 


3 


oo 


I 


oo 



Warren's Elements of Machine Construction and Drawing 8vo, 7 so 

Weisbach's Kinematics and the Power of Transmission. Herrmann — 

Klein.) 8vo, 5 00 

Machinery of Transmission and Governors. (Herrmann — Klein.). .8vo, 5 00 

HydrauLcs and Hydraulic Motors. (Du Bois.) 8vo, 5 00 

Wolff's Windmill as a Prime Mover 8vo, 3 00 

Wood's Turbines 8vo, 2 50 

MATERIALS OF ENGINEERING. 

Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edition, 

Reset 8vo, 7 50 

Church's Mechanics of Engineering 8vo, 6 00 

Johnson's Materials of Construction Large 8vo, 6 00 

Keep's Cast Iron 8vo 2 50 

Lanza's Applied Mechanics 8vo, 7 50 

Martens's Handbook on Testing Materials. (Henning.) 8vo, 7 50 

Merriman's Text-book on the Mechanics of Materials 8vo, 4 00 

Strength of Materials nmo, 1 00 

Metcalf's SteeL A Manual for Steel-users nmo. 2 00 

Smith's Wire : Its Use and Manufacture Small 4to, 3 00 

Materials of Machines i2mo, 1 00 

Thurston's Materials of Engineering 3 vols. , Svo 8 00 

Part H. — Iron and Steel 8vo, 3 SO 

Part IH. — A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents 8vo, 2 so 

Text-book of the Materials of Construction 8vo 5 00 

Wood's Treatise on the Resistance of Materials and an Appendix on the 

Preservation of Timber 8vo, 2 00 

Elements of Analytical Mechanics 8vo, 3 00 



STEAM-ENGINES AND BOILERS. 

Carnot's Reflections on the Motive Power of Heat. (Thurston.) 12 mo, 1 50 

Dawson's "Engineering" and Electric Traction Pocket-book. .i6mo, mor., 4 00 

Ford's Boiler Making for Boiler Makers i8mo, 1 00 

Goss's Locomotive Sparks 8vo, 2 00 

Hemenway's Indicator Practice and Steam-engine Economy i2mo, 2 00 

Hutton's Mechanical Engineering of Power Plants 8vo, 5 00 

Heat and Heat-engines 8vo, 5 00 

Kent's Steam-boiler Economy 8vo, 4 00 

Kneass's Practice and Theory of the Injector 8vo. 1 50 

MacCord's Slide-valves 8vo, 2 00 

Meyer's Modern Locomotive Construction 4to, 10 00 

Peabody's Manual of the Steam-engine Indicator i2mo, 1 50 

Tables of the Properties of Saturated Steam and Other Vapors 8vo, 1 00 

Thermodynamics of the Steam-engine and Other Heat-engines 8vo, 5 00 

Valve-gears for Steam-engines 8vo, 2 50 

Peabody and Miller's Steam-boilers 8vo, 4 00 

Pray's Twenty Years with the Indicator Large 8vo, 2 50 

Pupln's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. 

(Osterberg.) i2mo, 1 25 

Reagan's Locomotives : Simple, Compound, and Electric nmo, 2 50 

Rontgen's Principles of Thermodynamics. (Du Bois.) 8vo, 5 00 

Sinclair's Locomotive Engine Running and Management i2mo, 2 00 

Smart's Handbook of Engineering Laboratory Practice i2mo, 2 50 

Snow's Steam-boiler Practice 8vo, 3 00 

12 



Spangler's Valve-gears 8vo, 2 50 

Notes on Thermodynamics i2mo, 1 00 

Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 3 00 

Thurston's Handy Tables 8vo. 1 50 

Manual of the Steam-engine 2 vols.. 8vo 10 00 

Part I.— History, Structuce, and Theory 8vo, 6 00 

Part H. — Design, Construction, and Operation 8vo, 6 00 

Handbook of Engine and Boiler Trials, and the Use of the Indicator and 

the Prony Brake 8vo 5 00 

Stationary Steam-engines 8vo, 2 50 

Steam-boiler Explosions in Theory and in Practice i2mo, 1 50 

Manual of Steam-boilers , Their Designs, Construction, and Operation . 8vo, 5 00 

Weisbach's Heat, Steam, a id Steam-engines. (Du Bois.) 8vo, 5 00 

Whitham's Steam-engine I esign 8vo, 5 00 

Wilson's Treatise on Steam-boilers. (Flather.) i6mo, 2 50 

Wood's Thermodynamics. Heat Motors, and Refrigerating Machines 8vo, 4 00 



MECHANICS A.ND MACHINERY. 

Barr's Kinematics ot Machinery 8vo, 2^50 

Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Chase's The Art of Pattern-making i2mo, 2 S<> 

Chordal. — Extracts from Letters i2mo, 2 00 

Church's Mechanics of Engineering 8vo 6 00 

Notes and Examples in Mechanics 8vo. 2 00 

Compton's First Lessons in Metal-working 121110, 1 50 

Compton and De Groodt's The Speed Lathe i2mo, 1 50 

Cromwell's Treatise on Toothed Gearing i2mo, 1 50 

Treatise on Belts and Pulleys i2mo, 1 50 

Dana's Text-book of Elementary Mechanics for the Use of Colleges and 

Schools i2mo, 1 50 

Dingey's Machinery Pattern Making i2mo, 2 00 

Dredge's Record of the Transportation Exhibits Building of the World's 

Columbian Exposition of 1893 4to, half morocco, 5 00 

Du Bois's Elementary Principles of Mechanics : 

Vol. I. — Kinematics 8vo, 3 50 

Vol. H. — Statics 8vo, 4 00 

Vol. ni. — Kinetics 8vo, 3 50 

Mechanics of Engineering. Vol. I Small 4to, 7 50 

Vol. II Small 4to, 10 00 

Durley's Kinematics of Machines 8vo, 4 00 

Fitzgerald's Boston Machinist i6mo, 1 00 

Flather's Dynamometers, and the Measurement of Power i2mo, 3 00 

Rope Driving i2mo, 2 00 

Goss's Locomotive Sparks 8vo, 2 00 

Hall's Car Lubrication i2mo, 1 00 

Holly's Art of Saw Filing i8mo 75 

* Johnson's Theoretical Mechanics i2mo, 3 00 

Statics by Graphic and Algebraic Methods 8vo, 2 00 

Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 1 50 

Part H. — Form, Strength, and Proportions of Parts 8vo, 3 00 

Kerr's Power and Power Transmission 8vo, 2 00 

Lanza's Applied Mechanics 8vo, 7 50 

MacCord's Kinematics; or, Practical Mechanism 8vo, 5'' 00 

Velocity Diagrams. 8vo, 1 50 

Maurer's Technical Mechanics. (In preparation.} 

13 



4 


oo 


4 


00 


2 


50 


2 


00 


3 


00 


i 


50 


3 


00 


2 


00 


3 


00 


I 


00 


3 


00 


3 


00 


I 


00 


7 


50 


5 


00 


5 


00 


3 


00 


I 


25 


2 


50 


I 


00 



Merriman's Text-book on the Mechanics of Materials 8vo, 

* Michie's Elements of Analytical Mechanics 8sro, 

Reagan's Locomotives: Simple, Compound, and Electric i2mo, 

Reid's Course*in Mechanical Drawing 8vo, 

Text-book of^Mechanical Drawing and Elementary Machine Design. .8vo, 

Richards's Compressed Air i2mo, 

Robinson's Principles of Mechanism 8vo, 

Ryan, Norris, and Hoxie's Electrical Machinery. {In preparation.) 

Sinclair's Locomotive-engine Running andiManagement i2mo, 

Smith's Press-working of Metals 8vo, 

1 Materials of Machines i2mo, 

Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 

Thurston's Treatise on Friction and Lost Work in Machinery and Mill 
Work 8vo, 

Animal as a Machine and Prime Motor, and the Laws of Energetics. i2mo, 

Warren's Elements of Machine Construction and Drawing 8vo, 

Weisbach's Kinematics! and the Power of Transmission. (Herrmann — 
Klein.) 8vo, 

Machinery of Transmission and Governors. (Herrmann — Klein. ).8vo, 
Wood's Elements of Analytical Mechanics 8vo, 

Principles of Elementary Mechanics i2mo, 

Turbines 8vo, 

The World's Columbian Exposition of 1893 4to, 

METALLURGY. 

Egleston's Metallurgy of Silver, Gold, and Mercury: 

Vol. I. — Silver ... .8vo, 7 50 

Vol. II. — Gold and Mercury 8vo, 7 50 

** Iles's Lead-smelting. (Postage 9 cents additional.) nmo, 2 50 

Keep's Cast Iron 8vo, 2 50 

Kunhardt's Practice of Ore Dressing in Europe .8vo, 1 50 

Le Chatelier's High-temperature Measurements. (Boudouard — Burgess.) . i2mo, 3 00 

Metcalf 's Steel. A Manual for Steel-users i2mo, 2 00 

Smith's Materials of Machines nmo, 1 00 

Thurston's Materials of Engineering. In Three Parts 8vo, 8 00 

Part II. — Iron and Steel 8vo, 3 50 

Part III. — A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents . . .8vo, 2 50 

Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 

MINERALOGY. 

Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 

Boyd's Resources of Southwest Virginia 8vo, 

Map of Southwest Virginia Pocket-book form, 

Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 

Chester's Catalogue of Minerals 8vo, paper, 

Cloth, 

Dictionary of the Names of Minerals 8vo, 

Dana's System of Mineralogy Large 8vo, half leather, 

First Appendix to Dana's New "System of Mineralogy.". . . .Large 8vo, 

Text-book of Mineralogy 8vo, 

Minerals and How to Study Them i2mo, 

Catalogue of American Localities of Minerals Large 8vo, 

Manual of Mineralogy and Petrography i2mo, 

Egleston's Catalogue of Minerals and Synonyms 8vo, 

Hussak's The Determination of Rock-forming Minerals. (Smith.) Small 8vo, 

14 



2 


50 


3 


00 


2 


00 


4 


00 


1 


00 


1 


25 


3 


50 


12 


50 


1 


00 


4 


00 


1 


50 


1 


00 


2 


00 


2 


50 


2 


00 



* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo, paper, o 50 
Rosenbusch's Microscopical Physiography of the Rock-making Minerals. 

(Iddings.) 8vo, 5 00 

* Tillman's Text-book of Important Minerals and Docks 8vo, 2 00 

Williams's Manual of Lithology 8vo, 3 00 



MINING. 

Beard's Ventilation of Mines i2mo, 2 50 

Boyd's Resources of Southwest Virginia 8vo, 3 00 

Map of Southwest Virginia Pocket-book form, 2 00 

* Drinker's Tunneling, Explosive Compounds, and Rock Drills. 

4to, half morocco, 23 00 

Eissler's Modern High Explosives 8vo, 4 00 

Fowler's Sewage Works Analyses i2mo, 2 00 

Goodyear's Coal-mines of the Western Coast of the United States 121110, 2 50 

Dilseng's Manual of Mining 8vo, 4 00 

** Iles's Lead-smelting. (Postage 9c. additional.) i2mo, 2 50 

Kunhardt's Practice of Ore Dressing in Europe 8vo, 1 50 

O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 2 00 

* Walke's Lectures on Explosives 8vo, 4 00 

Wilson's Cyanide Processes i2mo, 1 50 

Chlorination Process i2mo, 1 50 

Hydraulic and Placer Mining i2mo, 2 00 

Treatise on Practical and Theoretical Mine Ventilation nmo, 1 25 



SANITARY SCIENCE. 

Copeland's Manual of Bacteriology. (In preparation.) 

FolwelTs Sewerage. (Designing, Construction, and Maintenance.; 8vo, 

Water-supply Engineering 8vo, 

Fuertes's Water and Public Health nmo, 

Water-filtration Works i2mo, 

Gerhard's Guide to Sanitary House-inspection i6mo, 

Goodrich's Economical Disposal of Town's Refuse Demy 8vo, 

Hazen's Filtration of Public Water-supplies 8vo, 

Kiersted's Sewage Disposal i2mo, 

Leach's The Inspection and Analysis of Food with Special Reference to State 

Control. (In preparation.) 
Mason's Water-supply. (Considered Principally from a Sanitary Stand- 
point.) 3d Edition, Rewritten 8vo, 

Examination of Water. (Chemical and Bacteriological.) i2mo, 

Merriman's Elements of Sanitary Engineering 8vo, 

Nichols's Water-supply. (Considered Mainly from a Chemical and Sanitary 

Standpoint.) (1883.) 8vo, 

Ogden's Sewer Design i2mo, 

* Price's Handbook on Sanitation i2mo, 

Richards's Cost of Food. A Study in Dietaries i2mo, 

Cost of Living as Modified by Sanitary^Science i2mo, 

Richards and Woodman's Air, Water, and Food from a Sanitary Stand- 
point 8vo, 

* Richards and Williams's The Dietary'Computer 8vo, 

Rideal's Sewage and Bacterial Purification of Sewage 8vo, 

Turneaure and Russell's Public Water-supplies 8vo, 

Whipple's Microscopy of Drinking-water 8vo, 

Woodhull's Notesland Military Hygiene i6mo, 

15 



3 


00 


4 


00 


I 


50 


2 


50 


I 


00 


3 


50 


3 


00 


I 


25 



4 


00 


I 


25 


2 


00 


2 


50 


2 


00 


I 


50 


I 


00 


I 


00 


2 


00 


I 


50 


3 


50 


5 


00 


3 


50 


1 


50 



MISCELLANEOUS. 

Barker's Deep-sea Soundings 8vo, 

Emmons's Geological Guide-book of the Rocky Mountain Excursion of the 

International Congress of Geologists Large 8vo, 

Ferrel's Popular Treatise on the Winds 8vo, 

Haines's American Railway Management nmo, 

Mott's Composition.'Digestibility , and Nutritive Value of Food. Mounted chart. 

Fallacy of the Present Theory of Sound i6mo, 

Ricketts's History of Rensselaer Polytechnic Institute, 1824- 1894. Small 8vo, 

Rotherham's Emphasized New Testament Large 8vo, 

Steel's Treatise on the Diseases of the Dog 8vo, 

Totten's Important Question in Metrology 8vo, 

The World's Columbian Exposition of 1893 4to, 

Worcester and Atkinson. Small Hospitals, Establishment and Maintenance, 
and Suggestions for Hospital Architecture, with Plans for a Small 
Hospital nmo, 1 25 

HEBREW AND CHALDEE TEXT-BOOKS. 

Green's Grammar of the Hebrew Language 8vo, 3 00 

Elementary Hebrew Grammar i2mo, 1 25 

Hebrew Chrestomathy 8vo, 2 00 

Gesenius's Hebrew and Chaldee Lexicon to the Old Testament Scriptures. 

(Tregelles.) Small 4to, half morocco, 5 00 

Letteris's Hebrew Bible 8vo, 2 25 

16 



I 


50 


4 


00 


2 


5o 


I 


25 


I 


00 


3 


00 


2 


00 


3 


50 


2 


5o 


1 


00 



NOV 13 1303 



itf-S 






■ 



H^H 






■ 






■ ■ 

■ 

■ I ^H Hi 

■ ■ 



m \ ■ 



■ 



■ • 






m &-i<zt*i L <*j 










■ € i 



